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CAJSWOW    SECTION 


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CANNON    SECTION 

STRESSES 

IN  WIRE-WRAPPED  GUNS 
AND    IN    GUN    CARRIAGES 


BY 
Lieutenant  Colonel   GOLDEN   L'H.    RUGGLES 

ORDNANCE  DEPARTMENT  U.  S.   ARMY 

Formerly  Professor  of  Ordnance  and  Science  of  Gunnery 
United  States  Military  Academy 


FIRST  EDITION 

FIRST   THOUSAND 


NEW   YORK 
JOHN   WILEY   &   SONS,    Inc. 

LONDON:    CHAPMAN    &    HALL,    LIMITED 
1916 


COPYRIGHT,  1916, 

BY 
LIEUT.  COL.  COLDEN   L'H.   RUGGLES 


Engwcriajt 


UF 

(*  O 


I? 


PREFACE  TO  SECOND  EDITION. 


This  text  was  originally  prepared  for  the  use  of  the  cadets  of 
the  United  States  Military  Academy  and  was  printed  by  the 
Military  Academy  Press.  On  this  account  it  has  not  hitherto 
been  available  to  the  public.  The  copies  of  the  original  edition 
being  exhausted,  a  new  edition  is  necessary  and  to  make  it  avail- 
able to  the  public  it  is  now  being  published  by  John  Wiley  &  Sons, 
Inc. 

The  author  desires  to  acknowledge  his  indebtedness  to  Lieut. 
Col.  Wm.  H.  Tschappat,  Ordnance  Department,  U.  S.  Army, 
Professor  of  Ordnance  and  Science  of  Gunnery,  U.  S.  Military 
Academy  for  correction  of  numerical  errors  in  the  original  text, 
and  for  arranging  for  the  present  publication;  to  Capt.  R.  H. 
Somers,  Ord.  Dept.,  for  correction  of  numerical  errors  and  proof 
reading;  and  to  Lieut.  J.  G.  Booton  C.A.C.,  and  T.  J.  Hayes,  4th 
Infantry,  for  proof  reading. 

GOLDEN     L'H.  RUGGLES. 
MAY  16, 1916. 


in 


523057 


PREFACE. 


In  this  text  the  author  has  endeavored  to  explain  and  illus- 
trate a  number  of  the  important  engineering  principles  underlying 
the  design  of  wire-wrapped  guns  and  of  gun  carriages,  and  in  its 
preparation  he  has  made  free  use  of  the  methods  of  officers  of 
the  Ordnance  Department,  U.  S.  Army,  who  have  been  engaged 
on  such  work  and  of  the  various  publications  on  the  subject 
written  by  them  or  translated  by  them  from  foreign  sources. 

The  deductions  of  the  formulas  relating  to  wire-wrapped  guns 
and  their  application  to  a  12-inch  gun  on  which  the  wire  is 
wrapped  under  constant  tension  have  been  taken  from  Notes  on 
the  Construction  of  Ordnance  Nos.  38  and  87  written  by  General 
William  Crozier,  Chief  of  Ordnance,  U.  S.  Army,  when  a  junior 
officer  of  the  Ordnance  Department,  the  formulas  giving  the 
tensions  in  the  wire  envelope  and  the  pressures  produced  by  it  be- 
ing, as  stated  by  General  Crozier,  mainly  those  of  Longridge,  an 
English  engineer.  Some  shortening  of  the  mathematical  work 
involved  in  the  deductions  of  the  formulas  has  been  effected  by 
the  author  of  this  text  by  starting  with  the  assumption  that  the 
modulus  of  elasticity  is  the  same  for  the  steel  wire  as  for  the 
steel  tube  of  the  gun,  and  the  formulas  have  been  slightly  ex- 
tended to  include  the  radial  stresses  in  the  wire  envelope. 

In  the  preparation  of  Chapter  II  much  assistance  was  derived 
from  the  ordnance  pamphlet  entitled  A  Discussion  of  the  Methods 
Proposed  to  Increase  the  Rapidity  of  Fire  of  Field  Guns  by  Captain 
(now  Lieutenant  Colonel)  Charles  B.  Wheeler  and  Captain  (now 
Major)  William  H.  Tschappat,  Ordnance  Department;  and  from 
the  original  calculations  made  in  the  Office  of  the  Chief  of  Ord- 
nance in  connection  with  the  design  of 'the  3-inch  field  carriage, 
model  of  1902,  and  the  5-inch  barbette  carriage,  model  of  1903. 

Much  assistance  was  likewise  derived  in  the  preparation  of 


Vi  PREFACE 

Chapter  III  from  the  original  calculations  made  in  the  Office  of  the 
Chief  of  Ordnance  in  connection  with  the  design  of  the  6-inch  dis- 
appearing carriage,  model  of  1905  Ml.  The  computations  re- 
lating to  the  throttling  grooves  of  this  carriage  given  in  the  text 
(which  are  the  same  in  principle  as  those  made  for  the  throttling 
grooves  of  a  number  of  earlier  models  of  disappearing  carriages) 
are  practically  identical  with  those  made  for  this  carriage  in  the 
Office  of  the  Chief  of  Ordnance  by  Captain  James  B.  Dillard, 
Ordnance  Department,  acting  under  the  direction  of  Major  John 
H.  Rice,  Ordnance  Department,  the  present  chief  of  the  gun- 
carriage  division  of  that  office. 

The  sources  of  the  formulas  used  in  Chapter  IV  are  given  in  the 
text. 

In  the  preparation  of  Chapter  V  the  author  has  freely  con- 
sulted various  standard  works  on  applied  mechanics  and  me- 
chanical engineering,  the  greatest  assistance  having  been  derived 
from  the  works  of  the  International  Library  of  Technology. 

The  methods  outlined  in  Chapter  VI  are  largely  based  upon  the 
practice  of  the  Ordnance  Department. 

The  thanks  of  the  author  are  due  to  Major  Tracy  C.  Dickson, 
Ordnance  Department,  for  advice  and  information  in  connection 
with  the  preparation  of  the  text;  and  to  Captain  Otho  V.  Kean, 
Ordnance  Department,  1st  Lieutenant  Ned  B.  Rehkopf,  2nd 
Field  Artillery,  and  1st  Lieutenant  George  R.  Allin,  6th  Field 
Artillery,  instructors  in  the  Department  of  Ordnance  and  Science 
of  Gunnery,  U.  S.  Military  Academy,  for  suggestions  tending  to 
add  to  the  clearness  of  the  text,  for  checking  and  correcting  where 
necessary  the  results  of  the  computations,  and  for  reading  the 
proofs. 

The  author  wishes  to  thank  also  Sergeant  Carl  A.  Schopper, 
Detachment  of  Ordnance,  U.  S.  Military  Academy,  for  the  skill 
and  care  with  which  he  has  prepared  the  many  drawings  for  the 
figures  appearing  in  the  text. 

GOLDEN  L'H.  RUGGLES. 
WEST  POINT,  NEW  YORK, 
May  25. 1910. 


CONTENTS. 


CHAPTER  I. 
Elastic  Strength  of  Wire- Wrapped  Guns.  —  Pages  1-36. 

General  construction,  1.  An  important  principle,  1.  Difference  between 
tangential  tension  and  tangential  stress  and  strain;  and  between  radial 
pressure  and  radial  stress  and  strain,  2.  The  elastic  strength  of  a  gun  is 
reached  when  the  stress  is  equal  to  the  elastic  limit  of  the  material;  Tension 
and  pressure  at  any  radius  of  a  compound  cylinder,  system  in  action  or  at 
rest,  4.  The  elastic  strength  of  a  compound  cylinder  properly  assembled  to 
secure  the  maximum  resistance  to  an  interior  pressure  depends  only  on  the 
sum  of  the  elastic  limits  for  compression  and  tension  of  the  material  of  the 
tube,  and  the  thickness  of  the  wall  in  calibers;  Compression  of  the  tube  at 
rest  beyond  its  elastic  limit,  5.  Two  principal  methods  of  wrapping  wire  on 
a  gun  tube;  Special  formulas  relating  to  layers  of  wire  wrapped  on  a  gun 
tube,  6. 

Case  I. — Wire  Wrapped  under  Constant  Tension,  8.  General  method  of 
design,  8.  Intensity  of  constant  tension  of  wrapping  necessary  to  produce 
a  given  pressure  on  the  exterior  of  the  tube,  system  at  rest,  9.  Intensity 
of  pressure  due  to  wrapping  at  any  radius  of  wire  envelope,  system  at  rest,  10. 
Intensity  of  tension  due  to  wrapping  at  any  radius  of  wire  envelope,  system 
at  rest,  11.  Tangential  and  radial  stresses  and  strains  due  to  wrapping  at  any 
radius  of  wire  envelope,  system  at  rest,  12.  Stresses  in  jacket,  wire  envelope, 
and  tube,  system  at  rest  and  in  action,  12. 

Example. — Section  through  the  Powder  Chamber  of  a  1 2-inch  Wire- Wrapped 
Gun  —  Wire  Wrapped  under  Constant  Tension,  13.  Maximum  permissible 
powder  pressure ;  Shrinkage  of  jacket ;  Constant  tension  of  wrapping,  14.  Stresses 
at  rest,  16.  Stresses  in  action  under  the  maximum  permissible  powder  pressure, 
19.  Stresses  in  action  under  a  powder  pressure  of  42000  Ibs.  per  sq.  in.,  22. 
Problem,  25. 

Case  II. — Wire  Wrapped  under  such  Varying  Tension  that  when  the  Gun 
is  Fired  with  the  Prescribed  Maximum  Powder  Pressure  all  Layers  of 
Wire  will  be  Subjected  to  the  Same  Tangential  Stress,  25.  General  method 
of  design,  25.  Radial  pressure  at  any  radius  of  wire  envelope,  system  in  action; 
Constant  tangential  stress  in  wire  envelope,  system  in  action,  26.  Tangential 
tension  and  radial  stress  at  any  radius  of  wire  envelope,  system  in  action,  27. 
Variable  tension  of  wrapping,  28.  Stresses  in  jacket,  wire  envelope,  and  tube, 
system  in  action  and  at  rest,  29. 

Example. — Section  through  the  Powder  Chamber  of  a  1 2-inch  Wire- 
Wrapped  Gun — Wire  Wrapped  under  Varying  Tension,  29.  Maximum 

vii 


viii  CONTENTS 

permissible  powder  pressure;  Pressure  on  exterior  of  tube  and  on  interior 
of  jacket,  system  in  action;  Shrinkage  of  jacket,  30.  Constant  tangential 
stress  in  wire  envelope,  system  in  action;  Variable  tension  of  wrapping,  30. 
Stresses  in  jacket  and  tube,  system  in  action  under  the  maximum  permissible 
powder  pressure,  and  system  at  rest,  31.  Stresses  m  wire  envelope,  system 
in  action  under  the  maximum  permissible  powder  pressure,  and  system  at 
rest,  32.  Stresses  in  action  under  a  powder  pressure  of  42000  Ibs.  per  sq.  in. 
34.  Problem,  36. 

CHAPTER  H. 

Determination  of  the  Forces  Brought  upon  the  Principal  Parts  of  the  3-Inch 
Field  Carriage  by  the  Discharge  of  the  Gun.  —  Pages  37-62. 

Stability;  Total  resistance  opposed  to  recoil  of  gun,  37.  Resultant  of 
forces  exerted  by  the  gun  on  the  carriage,  38.  Limiting  value  of  resistance 
to  recoil  compatible  with  stability  of  carriage,  38.  Method  of  varying  the 
resistance  to  recoil  followed  in  the  design  of  this  carriage,  42.  Relation 
between  the  varying  values  of  the  resistance  to  recoil  and  the  length  of  recoil, 
42.  Values  of  the  resistance  to  recoil  of  this  carriage,  44.  Margin  of  stability, 
45.  Velocity  of  restrained  recoil  as  a  function  of  space,  45. 

Forces  on  the  Carriage.  Forces  on  the  carriage  considered  as  one  piece,  46. 
Forces  on  the  gun,  48.  Forces  on  the  cradle,  50.  Forces  on  the  rocker,  52 
Forces  on  the  parts  below  the  rocker  considered  as  one  piece,  54. 

Problems. — Forces  on  the  5-inch  Barbette  Carriage,  Model  of  1903,  56. 
Forces  on  the  carriage  considered  as  one  piece,  57.  Forces  on  the  gun,  58. 
Forces  on  the  cradle,  59.  Forces  on  the  pivot  yoke,  60.  Forces  on  the 
pedestal,  62. 

CHAPTER  m. 

Determination  of  the  Forces  Brought  upon  the  Principal  Parts  of  a  Disappearing 
Gun  Carriage  by  the  Discharge  of  the  Gun.  —  Pages  63-105. 

The  6-inch  disappearing  carriage,  model  of  1905  Mi,  chosen  to  illustrate 
the  subject,  63.  Velocity  of  the  projectile  in  the  bore,  63.  Curve  of  recipro- 
cals; Velocity  and  travel  of  the  gun  in  free  recoil  while  the  projectile  is  in 
the  bore,  64.  Curve  of  free  recoil,  66.  Length  and  lettering  of  parts;  Reference 
to  co-ordinate  axes,  68.  Equations  expressing  the  relations  between  the 
forces  acting  on  the  gun,  70.  Equations  expressing  the  relations  between 
the  forces  acting  on  the  gun  levers,  72.  Equations  expressing  the  relations 
between  the  forces  acting  on  the  top  carriage,  74.  Equations  expressing  the 
relations  between  the  forces  acting  on  the  elevating  arm,  75.  Reduction  of 
the  number  of  unknown  quantities  in  these  equations,  76.  Method  of  deter- 
mining the  constant  resistance  of  the  recoil  cylinder,  77.  Determination  of 
the  values  of  the  co-ordinates  of  the  centers  of  mass  of  the  recoiling  parts  in 
terms  of  the  trigonometrical  functions  of  the  angles  which  the  parts  make 
with  the  axes  of  co-ordinates,  79.  Determination  of  the  values  at  any  time 
during  recoil  of  the  angle  ^  which  the  gun  makes  with  the  horizontal  axis 


CONTENTS  ix 

and  the  angle  9  which  the  elevating  arm  makes  with  the  vertical  axis,  when 
the  angle  $  which  the  lower  part  of  the  gun  levers  makes  with  the  vertical  axis 
is  known  or  assumed,  80.  Determination  of  the  linear  and  angular  velocities 
of  the  recoiling  parts  in  terms  of  the  trigonometrical  functions  of  the  angles 
and  of  d<j>/dt,  81.  Determination  of  the  linear  and  angular  accelerations  of 
the  recoiling  parts  in  terms  of  the  trigonometrical  functions  of  the  angles  and 
of  d?<l>/dP,  82. 

Determination  of  the  value  of  the  constant  resistance  of  the  recoil  cylinder  of 
this  carriage,  gun  fired  at  0  °  elevation,  85.  Determination  of  the  values  of  the 
forces  acting  on  the  carriage,  gun  fired  at  0°  elevation,  86.  Method  of  computing 
the  values  of  the  forces  when  the  gun  is  fired  at  any  angle  of  elevation,  90.  Maxi- 
mum values  of  the  forces  on  a  disappearing  carriage  during  recoil;  Velocity  and 
acceleration  curves  of  the  recoiling  parts,  91.  Method  of  computing  the  values  of 
the  forces  at  any  instant  while  the  projectile  is  in  the  bore,  92.  Method  of  com- 
puting the  values  of  the  forces  at  any  instant  after  the  powder  gases  have  ceased 
to  act  on  the  gun,  93.  Centripetal  accelerations,  93.  Effect  of  the  movement 
of  the  parts  on  the  intensities  of  the  forces,  94.  Formula  for  the  areas  of  the 
throttling  orifices  and  experimental  modification  to  allow  for  the  contraction  of 
the  liquid  vein,  95.  Profile  of  the  throttling  grooves,  97.  Velocities  of  the 
counterweight  and  the  top  carriage  compared,  101.  Velocity  and  acceleration 
curves  of  the  recoiling  parts  determined  before  the  carriage  is  built,  103. 
Method  of  designing  a  gun  carriage,  104. 

CHAPTER  IV. 
Stresses  in  Parts  of  Gun  Carriages.  —  Pages  106-173. 

Stresses  of  tension  and  compression,  106.  Shearing  stress,  107.  Torsional 
stress,  109.  Bending  stress,  111. 

Combined  Stresses,  118.  Tension  or  compression  and  bending,  118. 
Tension  or  compression  and  shear,  118.  Bending  and  shear,  119.  Shear 
and  torsion,  119.  Tension  or  compression  and  torsion,  120.  Bending  and 
torsion,  120.  Method  of  combining  stresses,  120. 

Neutral  Axis.     Center  of  Gravity.     Moment  of  Inertia,  121.     Neutral  axis, 

121.  Determination  of  centers  of  gravity  of  irregular  sections,  121.     Example, 

122.  Determination  of  moments  of  inertia  of  irregular  sections,  125.    Example, 
125.     Cubical  contents,  centers  of  gravity,  and  moments  of  inertia  of  irregu- 
lar volumes,  126.    Weight  and  mass  of  any  volume  of  a    given    material; 
Moment  of  inertia  of  the  mass  in  any  volume,  127. 

Permissible  stresses  in  gun-carriage  parts,  127. 

Stresses  in  Parts  of  the  3-inch  Field  Gun  and  Carriage,  Model  of  1902,  128. 
Stresses  in  the  front  clip  of  the  gun,  128.  Stresses  in  the  recoil  lug  of  the  gun, 
130.  Stresses  in  the  recoil  cylinder,  134.  Stresses  in  the  cradle,  137.  Stresses 
in  the  pintle  and  in  the  rivets  fastening  it  to  the  cradle,  143.  Stresses  in 
section  3-3  of  the  trail,  148. 

Stresses  in  Parts  of  the  s-inch  Barbette  Carriage,  Model  of  1903,  151. 
Stresses  in  the  trunnions  of  the  cradle,  153.  Stresses  in  the  recoil  cylinder 


X  CONTENTS 

due  to  the  interior  hydraulic  pressure,  151.  Stresses  in  section  4-4  of  the 
cradle,  153.  Stresses  in  section  5-5  of  the  pivot  yoke,  154.  Stresses  in  sec- 
tion 6-6  of  the  pivot  yoke,  159.  Stresses  in  section  7-7  of  the  pivot  yoke,  161. 
Stresses  in  section  8-8  of  the  pivot  yoke,  162.  Stresses  in  section  9-9  of  the 
pivot  yoke,  163.  Stresses  in  section  10-10  of  the  pedestal,  164.  Stresses  in 
the  foundation  bolts,  165.  Stresses  in  the  flange  at  the  rear  of  the  pedestal, 
165. 

Stresses  in  Parts  of  the  6-inch  Disappearing  Carriage,  Model  of  1905  Ml,  166. 
Section  11-11  of  the  gun  levers,  166.  Method  of  computing  stresses  in  parts 
not  in  equilibrium,  168.  Resolution  of  the  total  stresses  in  section  11-11  into 
unknown  horizontal  and  vertical  components  Ps  and  Pt  and  an  unknown 
couple  YY,  169.  Determination  of  the  values  of  Ps,  Pt,  and  YY,  169.  Shearing 
stress,  tensile  stress,  and  bending  stress  in  section  11-11,  171.  Maximum 
resultant  stress  in  section  11-11, 172.  Stresses  in  the  trunnions  of  the  gun-lever 
axle,  172.  Stresses  in  the  elevating  arm  and  other  parts  of  this  carriage,  173. 

CHAPTER  V. 
Toothed  Gearing.  —  Pages  174-225. 

Definition;  Ratio  of  angular  velocities,  174.  Necessity  for  teeth;  Tooth 
surfaces  must  have  certain  definite  forms,  175.  Outline  of  gear-teeth;  Pitch 
circumference;  Circular  pitch;  Diametral  pitch;  176.  Angular  velocity  of 
rotating  wheel,  177.  Condition  to  be  fulfilled  by  tooth  curves,  179.  The 
involute  to  a  circle,  180.  The  involute  system  of  gear-teeth,  181.  The 
cycloid;  Epicycloid;  Hypocycloid,  184.  The  cycloidal  system  of  gear-teeth, 
185.  Spur  gears,  188.  Rack  and  spur  gear,  190.  Bevel  gears,  190.  Screw 
gearing;  Worm  and  worm-wheel,  191.  Spiral  gears,  198.  Distinction 
between  worm-gearing  and  spiral  gearing,  200.  Velocity  ratio  of  worm- 
gears,  201.  Velocity  ratio  of  spiral  gears,  202.  Tooth  curves  of  screw  gears, 
202.  Shafts  connected  by  screw  gearing;  Drivers;  Pitch  of  screw  gearing, 
202.  Wheel  train;  Velocity  ratio  of  first  and  last  shafts  connected  by  a 
wheel  train,  204.  Idlers,  205.  Relation  between  power  and  resistance  in  a 
wheel  train  (a)  friction  neglected,  207,  (6)  friction  of  gearing  considered, 
208.  Efficiencies  of  various  classes  of  gears,  209. 

Example.  Force  on  the  crank-handle  of  the  5-inch  barbette  carriage, 
model  of  1903,  required  to  start  the  gun  and  to  keep  it  moving  in  elevation 
with  a  uniform  angular  velocity,  210.  Force  on  the  crank-handle  required  to 
produce  in  a  given  time  a  given  angular  velocity  of  the  gun  in  elevation,  212. 

Pressure  between  the  teeth  of  any  pair  of  gears  in  a  wheel  train,  213. 
Stresses  in  the  shafts,  214.  Stresses  in  the  gear-teeth,  215.  Stresses  in  the 
arms  of  gear-wheels,  216.  Proportions  for  the  rims  of  gear-wheels,  216. 
Proportions  for  the  hubs  of  gear-wheels,  217. 

Gear-Cutting,  217.  Rotary  cutters,  217.  Cutting  the  teeth  of  spur 
gears,  218.  Cutting  the  teeth  of  spiral  gears,  219.  Cutting  the  teeth  of  a 
worm  and  worm-wheel,  219.  Cutting  the  teeth  of  bevel  gears  with  a  rotary 
cutter,  221.  Planing  the  teeth  of  bevel  gears,  225. 


CONTENTS  XI 

CHAPTER  VI. 
Counter-Recoil  Springs.  —  Pages  228-259. 

Helical  springs,  228.  Torsional  stresses  and  strains  in  a  straight  bar,  228. 
Stresses  in  helical  springs,  231.  Fundamental  equations  relating  to  helical 
springs,  233.  Helical  springs  coiled  from  bars  of  circular  cross-section,  236. 
Helical  springs  coiled  from  bars  of  rectangular  cross-section,  237.  Require- 
ments to  be  fulfilled  by  a  counter-recoil  spring;  Nomenclature,  238.  Design 
of  counter-recoil  springs  coiled  from  bars  of  circular  cross-section,  239.  Design 
of  counter-recoil  springs  coiled  from  bars  of  rectangular  cross-section,  244. 
Telescoping  springs,  246.  Design  of  telescoping  springs,  247.  Design  of 
non-telescoping  springs  assembled  one  within  the  other,  249.  Measures  for 
decreasing  assembled  height  applicable  either  to  springs  coiled  from  bars  of 
circular  cross-section  or  to  those  coiled  from  bars  of  rectangular  cross-sec- 
tion, 251.  Counter-recoil  springs  for  the  5-inch  barbette  carriage,  model  of 
1903,  251. 

Examples.  Design  of  counter-recoil  springs  coiled  from  bars  of  rectangular 
cross-section  for  use  in  the  spring  cylinder  of  the  5-inch  barbette  carriage, 
model  of  1903,  252.  Assembled  height  of  a  column  of  counter-recoil  springs 
for  this  carriage  coiled  from  bars  of  circular  cross-section,  254.  Assembled 
height  of  a  column  of  counter-recoil  springs  for  this  carriage  formed  by  placing 
springs  of  smaller  diameter  inside  the  larger  springs,  both  sets  of  springs 
being  coiled  from  bars  of  circular  cross-section  and  acting  directly  on  the 
spring  piston,  255.  Design  of  telescoping  springs  coiled  from  bars  of  circular 
cross-section  for  use  in  this  carriage  without  alteration  of  the  spring  cylinder, 
256.  Assembled  height  of  a  column  of  telescoping  springs  for  this  carriage 
coiled  from  bars  of  circular  cross-section  determined  under  the  assumption 
that  the  spring  cylinder  may  be  altered  to  permit  the  ends  of  the  inner  spring 
column  and  the  stirrup  to  be  drawn  through  an  opening  in  the  cylinder 
during  recoil,  258. 


BN8INEERIN0 

GAUKCtf 


Stresses  in  Wire- Wrapped  Guns  and  in 
Gun  Carriages. 


CHAPTER  I. 
ELASTIC  STRENGTH  OF  WIRE-WRAPPED  GUNS. 

1.  General     Construction.  —  Wire- wrapped    guns    consist    of 
(a)  an  inner  steel  tube  which  forms  the  support  on  which  the 
wire  is  wrapped  and  in  which  the  rifling  grooves  are  cut;   (b)  the 
layers  of  wire  wrapped  upon  the  tube  to  increase  its  resistance 
by  the  application  of  an  exterior  pressure  as  well  as  to  add  to  the 
strength  of  the  structure  by  their  own  resistance  to  extension 
under  fire;  and  (c)  one  or  more  layers  consisting  of  a  steel  jacket 
and  hoops  placed  over  the  wire  with  or  without  shrinkage.     The 
jacket  is  generally  relied  upon  to  furnish  the  longitudinal  strength 
of  the  gun  since  this  feature  is  lacking  in  the  wire  envelope.     The 
breech-block  is  therefore  ordinarily  screwed  into  the  jacket,  or 
into  a  breech  bushing  screwed  into  the  jacket,  in  order  that  the 
latter  may  resist  the  longitudinal  stress  due  to  the  pressure  at 
the  bottom  of  the  bore,  the  consequent  rearward  acceleration  of 
the  gun  in  recoil,  and  the  resistance  to  recoil  of  the  recoil  brake 
applied  to  the  gun  through  the  trunnions  or  through  a  lug  form- 
ing part  of  the  jacket. 

2.  An  Important  Principle.  —  A  very  important   principle  in 
wire   gun   construction   is   that   enunciated   in   paragraph    121, 
page  216,  Lissak's  Ordnance  and  Gunnery,  to  the  effect  that 
the  stresses  produced  by  any  pressure  applied  to  a  compound 
cylinder  are  exactly  the  same  as  would  be  produced  by  the  same 
pressure  applied  to  a  single  cylinder  of  the  same  dimensions. 
This  refers  to  the  stresses  and  strains  produced  by  the  pressure 
under  consideration  only,  and  if  before  the  application  of  this 
pressure  there  existed  in  the  compound  cylinder  stresses  and 

1 


2  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

strains  produced  by  shrinkage  or  otherwise,  the  resulting  stresses 
are  the  algebraic  sum  of  those  previously  existing  and  those 
induced  by  the  application  of  the  pressure.  It  follows  from  this 
that  if  we  know  the  tangential  and  radial  stresses  and  strains 
existing  at  any  radius  r  in  a  compound  cylinder  before  the  appli- 
cation of  an  interior  or  exterior  pressure,  the  resultant  tangential 
and  radial  stresses  and  strains  therein  at  any  radius  r  when  an 
interior  or  exterior  pressure  or  both  are  acting  can  be  readily 
obtained  by  adding  algebraically  to  those  previously  existing 
the  stresses  and  strains  computed  from  the  appropriate  one  of 
equations  (9)  and  (10),  paragraph  104,  page  195,  Lissak's  Ord- 
nance and  Gunnery,  which  are  as  follows: 


O    D   Z?  2          P  Z?  2         /I    Z?   Z?  2  (T>  T)  \ 

Elt    =    St=\    P°g°2     ~  g™     +  I  RJ^R,  I/?"2        '(1) 

_  2  Ppflo2  -  P^2      4  R0*Rt  (P0  -  PQ 

£/tp    —    Op    —    Q    p-£ p~2 ~~   Q pTTf ~p-£ 1/7™         (6) 

O          /Vl  ito  <5  -ftl     —   "0 

in  which  R0  and  J?i  are  the  inner  and  outer  radii,  respectively, 
of  the  compound  cylinder  and  r  is  the  radius  of  any  point  within 
the  wall  of  the  cylinder.  If  the  compound  cylinder  is  composed 
of  any  number  of  layers  n  and  its  radii  are  designated  as  R0, 
Ri,  R2,  .  .  .  Rn,  Ro  in  formulas  (1)  and  (2)  will  be  the  R0  of  the 
compound  cylinder  but  the  Ri  of  these  formulas  will  be  the  Rn  of 
the  compound  cylinder,  r  in  the  formulas  may  have  any  value 
between  R0  and  Rn  of  the  compound  cylinder. 

Similarly  if  we  know  the  resultant  tangential  and  radial  stresses 
and  strains  at  any  radius  r  in  a  compound  cylinder  when  acted 
upon  by  an  interior  or  exterior  pressure,  or  by  both,  the  tan- 
gential and  radial  stresses  and  strains  that  will  remain  at  any 
radius  r  when  these  pressures  are  removed  can  be  readily  obtained 
by  subtracting  algebraically  from  the  resultant  stresses  and 
strains  those  computed  from  the  appropriate  one  of  equations 
(1)  and  (2)  as  due  to  the  pressures  that  have  been  removed. 

3.  Difference  Between  Tangential  Tension  (or  Tension  Simply) 
and  Tangential  Stress  and  Strain ;  and  Between  Radial  Pressure 
and  Radial  Stress  and  Strain.  —  Equations  (7)  and  (8),  page  194, 
Lissak's  Ordnance  and  Gunnery,  which  are  as  follows: 

*  The  equations  in  this  discussion  will  be  renumbered  consecutively  as  used 
regardless  of  their  number  in  Lissak's  Ordnance  and  Gunnery. 


ELASTIC  STRENGTH  OF  WIRE-WRAPPED  GUNS 


_      po2  -  PJt?   ,  floW  (P,  -  PQ  , 
2  -       2  2  -       2 


oo    -  fifr2       flpW  (P0  -  PQ 
flx2  -  #o2  #i2  -  #o2 

differ  from  equations  (1)  and  (2)  in  that  they  give  the  tangential 
tension  t  and  the  radial  pressure  p  at  any  radius  r  produced  by 
the  application  of  interior  and  exterior  pressures  to  a  cylinder 
whose  inner  and  outer  radii  are  R0  and  Ri,  respectively.  In  the 
deduction  of  these  equations  tensile  forces  and  tensile  strains 
have  been  considered  positive  and  compressive  forces  and  com- 
pressive  strains  negative.  The  radial  pressure  in  the  walls  of  a 
gun  is  always  a  compressive  force.  The  difference  between  the 
tension  and  the  tangential  stress  and  strain,  and  between  the 
radial  pressure  and  the  radial  stress  and  strain  should  be  carefully 
noted.  The  tangential  strain  lt  is  due  both  to  the  tangential 
tension  (or  compression)  and  the  radial  pressure  and  is  equal  to 

l/tf[*-(-p/3)]     or     l/E(t  +  p/S) 

obtained  from  the  first  of  equations  (4),  page  192,  Lissak's  Ord- 
nance and  Gunnery,  by  making  q  =  0;  and  the  tangential  stress 
Elt  is  equal  to  this  strain  multiplied  by  the  modulus  of  elasticity 
E,  or  to  (t  +  p/3).  If  the  tension  and  the  pressure  have  differ- 
ent signs,  that  is,  if  one  is  a  tensile  and  the  other  a  compressive 
force,  the  tangential  stress  will  be  greater  numerically  than  the 
tension  by  one-third  of  the  radial  pressure;  but  if  they  have  like 
signs,  that  is,  if  both  are  tensile  or  compressive  forces  the  tan- 
gential stress  will  be  equal  numerically  to  the  difference  between 
the  tangential  tension  and  one-third  of  the  radial  pressure. 
Similarly  the  radial  strain  lp  is  due  both  to  the  tangential  tension 
and  the  radial  pressure  and  is  equal  to 

l/E  (-p  -  «/3)  =  -\/E  (p  +  f/3) 

obtained  from  the  second  of  equations  (4),  page  192,  Lissak's 
Ordnance  and  Gunnery,  by  making  q  =  0;  and  the  radial  stress 
Elp  is  equal  to  this  strain  multiplied  by  the  modulus  of  elasticity 
E,  or  to  --  (p  +  £/3).  If  the  tension  and  the  pressure  have 
different  signs  the  radial  stress  will  be  numerically  greater  than 
the  radial  pressure  by  one-third  of  the  tangential  tension;  but 
if  they  have  like  signs  the  radial  stress  will  be  equal  numerically 


4  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

to  the  difference  between  the  radial  pressure  and  one-third  of 
the  tangential  tension. 

4.  The  Elastic  Strength  of  a  Gun  is  Reached  when  Either  /•;/. 
or  Elp  is  Equal  to  the  Elastic  Limit  of  the  Material.  —  Tension 
and  Pressure  at  any  Radius  r  of  a  Compound  Cylinder,  System 
in  Action  or  at  Rest.  —  While  as  a  matter  of  fact,  neglecting  the 
longitudinal  force,  the  only  forces  acting  on  a  particle  in  the  walls 
of  a  gun  are  the  tangential  tension  and  the  radial  pressure,  the 
strain  produced  by  these  forces  acting  together  is,  in  the  tan- 
gential or  radial  direction,  the  same  as  would  be  produced  by  a 
force  Elt  or  Elp,  respectively,  acting  alone.     It  is  generally  ac- 
cepted that  if  the  material  is  subjected  to  a  compressive  or  ten- 
sile strain  equal  to  that  occurring  at  the  elastic  limit  when  a  single 
force  is  applied  in  the  direction  of  the  strain,  it  will  yield  no  matter 
how  the  strain  may  be  produced;   and  therefore  the  stresses  Elt 
and  Elp  corresponding  to  the  strains  lt  and  lp  instead  of  the  tan- 
gential tension  and  the  radial  pressure  are  considered  as  deter- 
mining whether  or  not  the  metal  of  the  gun  is  being  worked  within 
the  elastic  limit. 

As  in  the  case  of  the  stresses  and  strains,  the  tensions  and 
pressures  produced  by  any  interior  or  exterior  pressure  or  both 
applied  to  a  compound  cylinder  are  exactly  the  same  as  would 
be  produced  by  the  same  pressure  or  pressures  applied  to  a  single 
cylinder  of  the  same  dimensions,  and  if  we  know  the  tension  and 
pressure  at  any  radius  r  of  the  compound  cylinder  before  the 
application  of  the  pressures,  the  resultant  tension  and  pressure 
at  that  radius  when  the  pressures  are  acting  may  be  determined 
by  adding  algebraically  to  those  previously  existing,  those  due 
only  to  these  pressures,  computed  from  equations  (3)  and  (4). 
Also  if  we  know  the  resultant  tension  and  pressure  which  exist 
at  any  radius  r  of  a  compound  cylinder  when  interior  or  exterior 
pressures  are  acting,  those  which  will  remain  at  the  radius  r  when 
the  interior  or  exterior  pressures  are  removed  can  be  obtained 
by  subtracting  algebraically  from  the  resultant  tension  and 
pressure  those  computed  by  equations  (3)  and  (4)  as  due  only 
to  the  pressures  which  have  been  removed. 

5.  The  Elastic  Strength  of  a  Compound   Cylinder,  Properly 
Assembled  to  Secure  the  Maximum  Resistance  to  an  Interior 
Pressure,  Depends  only  on  the  Sum  of  the  Elastic  Limits  for 


ELASTIC  STRENGTH  OF  WIRE-WRAPPED  GUNS  5 

Compression  and  Tension  of  the  Material  of  the  Tube  and  the 
Thickness  of  the  Wall  in  Calibers.  —  Compression  of  the  Tube, 
System  at  Rest,  Beyond  its  Elastic  Limit.  —  It  follows  from 
the  above  discussion  that  if  a  compound  cylinder  has  been  so 
assembled  as  to  compress  the  inner  surface  of  the  inner  cylinder 
to  the  elastic  limit  p  for  that  cylinder,  the  elastic  strength  of  the 
compound  cylinder  can  be  determined  from  equation  (1)  by  find- 
ing that  interior  pressure  which  acting  alone  on  the  compound 
cylinder  will  cause  a  stress  p  +  0  on  its  interior  surface,  which 
stress  will  overcome  the  initial  compression  p  and  cause  a  tan- 
gential extension  equal  to  6,  the  elastic  limit  for  tension  of  the 
material. 

Making  PI  =  Pn  =  0,  Ri  =  Rn,  r  =  R0,  and  St  =  P  +  9  in 
equation  (1)  and  solving  for  P0,  we  have 


n  -  flo2)  (P  +  0) 
22 


which  gives  the  maximum  interior  pressure  which  the  compound 
cylinder  can  withstand  without  exceeding  the  elastic  limit  for 
tension  of  the  inner  cylinder.  In  this  discussion  it  is  assumed 
that  the  compound  cylinder  whether  wire-wrapped  or  of  the  built- 
up  construction  has  been  properly  assembled,  in  which  case  the 
stresses  in  the  layers  outside  the  inner  cylinder  will  not  exceed 
the  elastic  limit  whether  in  the  state  of  rest  or  action.  As  shown 
in  paragraph  121,  page  217,  Lissak's  Ordnance  and  Gunnery, 
the  greatest  value  of  P0,  corresponding  to  Rn  =  oo  ,  is  .75  (p  +  0). 
Making  Rn  =  3  RQ,  corresponding  to  a  thickness  of  wall  of  one 

caliber,  ' 

Po  =  .63  (p  +  6) 

and,  therefore,  comparatively  little  advantage  results  from  in- 
creasing the  thickness  of  wall  beyond  one  caliber.  On  the  other 
hand,  whatever  be  the  mode  of  assembling  the  compound  cylinder, 
by  wrapping  the  tube  with  wire  or  otherwise,  the  elastic  strength 
of  the  cylinder,  if  properly  assembled  to  secure  the  maximum 
resistance  to  an  interior  pressure,  will  depend  only  on  the  sum  of 
the  elastic  limits  for  compression  and  tension  of  the  tube  and  the 
thickness  of  the  wall  in  calibers. 

Many  designers  of  wire-wrapped  guns  have  compressed  the 
tube  in  the  state  of  rest  beyond  the  elastic  limit  for  tangential 


6 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


compression,  and  on  the  assumption  that  the  elastic  limit  for 
tension  has  not  been  lowered  thereby,  have  computed  the  elastic 
strength  of  the  gun  from  equation  (1),  substituting  for  p  the 
value  of  the  actual  stress  of  compression  of  the  bore  of  the  tube. 
Many  experiments,  however,  indicate  that  if  the  metal  is  com- 
pressed beyond  the  elastic  limit,  its  elastic  limit  for  tension  is 
lowered,  and  vice  versa.  Therefore,  if,  when  such  a  gun  is  fired, 
the  stress  at  the  bore  is  raised  to  the  elastic  limit  for  tension  0 
as  determined  in  the  testing  machine,  it  will  in  reality  have 
passed  the  actual  elastic  limit,  due  to  the  over-compression  at 
rest.  As  a  result  the  tube  will  be  worked  beyond  its  elastic 
limit  both  in  tension  and  compression.  Such  treatment  is  known 
to  be  most  injurious  if  the  repetitions  of  stress  are  sufficiently 
numerous,  but  owing  to  the  comparatively  limited  number  of 
rounds  fired  in  any  gun  no  trouble  has  so  far  resulted  in  wire- 
wrapped  guns  from  over-compression  of  the  tube. 

6.  Two  Principal  Methods  of  Wrapping  Wire  on  a  Gun  Tube. 
Special  Formulas  Relating  to  Layers  of  Wire  Wrapped  on  a  Gun 
Tube.  —  Let  Fig.  1  represent  a  section  of  a  wire-wrapped  gun 


Fig.  1. 

consisting  of  tube,  wire  envelope,  and  jacket.  Represent  the 
inner  and  outer  radii  of  the  tube  by  RQ  and  Ri,  respectively,  the 
radius  of  the  outer  surface  of  jthe  wire  envelope  by  R%  and  that  of 
the  outer  surface  of  the  jacket  by  R3.  When  wire  is  wrapped 
under  tension  around  a  tube  the  effect  is  to  cause  a  pressure  on 
its  exterior  surface  which  compresses  the  metal  tangentially,  the 
greatest  stress  occurring  as  has  been  shown  (paragraph  108,  page 
200,  Lissak's  Ordnance  and  Gunnery)  at  the  bore.  The  pressure 
on  the  tube  and  the  resulting  compression  in  a  tangential  direction 
increase  with  the  number  of  layers  of  wire  applied.  As  each 


ELASTIC  STRENGTH  OF  WIRE-WRAPPED  GUNS  7 

layer  of  wire  is  applied  its  tension  is  that  of  wrapping,  but  this  is 
gradually  reduced  by  the  compression  of  this  layer  due  to  the 
pressure  on  its  surface  caused  by  the  application  of  the  succeeding 
layers  of  wire.  There  are  two  principal  methods  of  wrapping 
wire  on  a  tube;  one  is  to  wrap  the  wire  at  constant  tension  and 
the  other  is  to  wrap  it  at  such  varying  tension  that,  when  the 
gun  is  fired  with  the  prescribed  pressure,  all  the  layers  of  wire 
shall  be  subjected  to  the  same  tangential  stress.  The  latter 
method  is  theoretically  the  better,  but  owing  to  the  greater 
convenience  of  wrapping  the  wire  at  constant  tension  and  to  its 
great  elastic  strength,  which  permits  the  tube  to  be  compressed 
to  its  elastic  limit  by  wrapping  wire  thereon  at  constant  tension 
without  causing  the  stress  in  the  wire  to  approach  too  close  to 
its  elastic  limit  when  the  gun  is  fired,  the  former  method  is  gen- 
erally used.  Formulas  have  been  deduced  giving 

(a)  the  uniform  tension  T  at  which  an  envelope  of  wire  of  thick- 
ness equal  to  Rz  —  R\  must  be  wrapped  on  a  tube  of  radii 
R0  and  Ri  to  produce  a  pressure  P/  on  its  exterior. 

(6)  the  pressure  at  any  radius  r  of  the  envelope  of  wire  when  the 
gun  is  in  the  state  of  rest  due  to  the  wrapping  thereon  of 
the  outer  layers  of  wire. 

(c)  the  resultant  tension  at  any  radius  r  of  the  envelope  of  wire 

due  to  the  tension  of  wrapping  and  to  the  compression 
caused  by  the  wrapping  of  the  layers  from  that  radius  out. 

(d)  the  tangential  and  radial  stresses  and  strains  at  any  radius  r 

of  the  wire  envelope  after  the  wrapping  has  been  completed. 

(e)  the  uniform  tangential  stress  Elt  =  St  which  must  occur 

in  all  the  layers  of  a  wire  envelope  of  thickness  equal  to 
R2  —  Ri  to  produce  a  pressure  PI  on  the  exterior  of  a  tube 
of  radii  R0  and  Ri  when  the  gun  is  fired. 

(/)  the  variable  tension  of  winding  which  must  be  used  in  order 
that  when  the  gun  is  fired  all  the  layers  of  wire  will  be  sub- 
jected to  a  uniform  tangential  stress  St* 

(0)  the  pressure,  the  tension,  and  the  radial  stress  and  strain 
which  exist  at  any  radius  r  of  the  wire  envelope  when  the 
gun  is  fired. 

The  formulas  referred  to  under  (a)  to  (d),  inclusive,  pertain 
only  to  the  case  where  the  wire  is  wrapped  under  constant  ten- 


8  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

sion  and  relate  exclusively  to  the  gun  in  the  state  of  rest.  Those 
referred  to  under  (e)  to  (</),  inclusive,  pertain  only  to  the  case 
where  the  wire  is  wrapped  at  such  varying  tension  that  when  the 
gun  is  fired  with  the  prescribed  powder  pressure  all  layers  of 
wire  shall  be  subjected  to  the  same  tangential  stress;  and  they 
relate  exclusively  to  the  gun  in  the  state  of  action. 

CASE  I.     WIRE  WRAPPED   UNDER  CONSTANT  TENSION. 

7.  General  Method  of  Design.  —  Suppose  it  is  desired  to  con- 
struct a  wire-wrapped  gun  as  shown  in  Fig.  1,  in  such  manner 
as  to  give  to  it  the  maximum  elastic  strength  permitted  by  the 
quality  of  the  metal  in  the  tube  and  the  total  thickness  of  wall 
#3  —  RQ,  with  the  condition  that  the  wire  shall  be  wrapped  on 
the  tube  under  constant  tension.  The  first  thing  to  determine 
is  the  maximum  interior  pressure  which  the  gun  can  support  in 
action  on  the  assumption  that  in  the  state  of  rest  the  inner  sur- 
face of  the  tube  is  compressed  to  its  elastic  limit.  This  is  given 
by  equation  (5).  The  exterior  pressure  on  the  tube  which  will 
cause  its  interior  surface  to  be  compressed  to  the  elastic  limit 
when  in  the  state  of  rest  is  then  determined  from  equation  (29), 
page  206,  Lissak's  Ordnance  and  Gunnery,  which,  after  replacing 
a  by  its  value  Ri2/R<?  is  as  follows: 

D  ^!2    ~~   ^<>2  /£\ 

Pl"=  p 


Part  of  this  pressure  Pip  will  be  due  to  the  wire  envelope  and  part 
to  the  shrinkage  of  the  jacket  on  the  wire.  If  the  jacket  is 
assembled  without  shrinkage,  as  is  sometimes  the  case,  all  of  the 
pressure  P\p  will  be  due  to  the  pressure  exerted  by  the  wire  en- 
velope. Supposing  the  jacket  assembled  with  shrinkage,  the 
amount  of  the  shrinkage  must  be  determined  and  from  that  the 
pressure  on  the  exterior  of  the  tube  due  to  such  shrinkage.  Sub- 
tracting this  pressure  from  Pip  there  results  the  pressure  which 
must  be  exerted  by  the  wire  envelope.  To  utilize  the  full  strength 
of  the  jacket  its  inner  surface  must  be  extended  in  the  state  of 
action  to  the  elastic  limit.  Part  of  this  extension  will  be  due  to 
the  action  of  P0  and  the  remainder  to  the  shrinkage.  The  part 
due  to  Po  is  obtained  from  equation  (1)  by  making  Pi  =  0,  Ri  = 
R3,  and  r  =  R2.  The  difference  between  this  part  and  the  tensile 


ELASTIC  STRENGTH  OF  WIRE-WRAPPED  GUNS    .  9 

elastic  limit  03  is  due  to  the  shrinkage.  Calling  this  difference 
S't3,  the  pressure  P'2  to  produce  it  is  obtained  from  equation  (1) 
by  making  Pl  =  0,  St  =  S't3,  P0  =  P'2>  fli  =  R3,  R0  =  R2,  and 
r  =  R2,  and  solving  for  P'2,  whence 

3  (fl32  -  R      ?,  ~ 

'3 


The  absolute  shrinkage  to  produce  this  pressure  is  given  by 
equation  (54),  page  218,  Lissak's  Ordnance  and  Gunnery,  which, 
after  replacement  of  the  radius  ratios  by  their  values  in  terms  of 
radii  and  making  Ri  =  R%,  R2  —  Ra,  Pit  =  P'2,  and  Si  =  82, 
becomes  * 

(fl32-flo2)P'2  ,R, 

2          -     2 


*  ~  E  (R2*  -  #o2)  (Rt  -  #22) 

The  pressure  P'2  due  to  the  shrinkage  of  the  'jacket  on  the  wire 
envelope  causes  a  pressure  p'\  on  the  exterior  of  the  tube  which  is 
obtained  from  equation  (4)  by  making  P0  =  0,  PI  =  P'2,  Ri  =  R2, 
and  r  =  Ri. 
Whence 


P'tRS     L  _  Rl 
RS-  R<?  V1      RS 


The  pressure  p\  to  be  produced  by  the  wire  envelope  is,  there- 
fore, 

p\  =  Plp  -  p'\ 

8.  Intensity  of  the  Constant  Tension  of  Wrapping  Necessary 
to  Produce  a  Given  Pressure  on  the  Exterior  of  the  Tube,  System 
at  Rest.  —  It  now  remains  to  deduce  the  formulas  referred  to 
under  (a)  to  (d),  inclusive,  article  6.  To  do  this  let  us  suppose 
the  wrapping  of  the  wire  on  the  tube,  Fig.  1,  to  be  finished  to  a 
radius  r  producing  a  pressure  on  the  exterior  of  the  tube  pt  and 
then  to  be  continued  to  any  other  radius  r',  the  pressure  on  the 
exterior  of  the  tube  changing  to  p'  t;  the  additional  wrapping  will 
also  produce  a  pressure  pr  on  the  wire  at  r.  The  pressure  pr  on 
the  wire  at  r  then  produces  a  pressure  p't  —  Pt  on  the  tube. 
Applying  equation  (4),  making  p  =  p't  —  Pt,  P0  =  0,  PI  =  pr> 
r  =  Ri,  Ri  —  r,  we  obtain 


10        ,        STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

which,  by  reduction  becomes, 

_  (Rf-Rfirtpr 
Pt      P'-   RS(r*-Rft 

Let  r  be  the  mean  tension  of  the  wires  between  r  and  r',  then 

pr  X  2  r  =  T  X  2  (r1  —  r}  whence 

r'  -r 


Substituting  this  value  for  pr  in  equation  (11)  and  dividing  by 
r'  —  r,  we  get 

p'»  -pt  =  (R,2  -  R02)  rr 

r*  -  r       RS  (r2  -  R<?) 

Passing  to  the  limit  of  both  members,  we  have,  since  T  then  be- 
comes the  uniform  tension  of  wrapping  T, 

(ft  2   ft  2~\  rm 

r>  2  /.v.2 r>  2\  \-*-"/ 


Multiplying  by  dr  and  integrating  between  the  limits  r  and 

2  -      2  *2  -  R<? 


which  gives  the  pressure  on  the  tube  produced  by  wrapping  from 
r  to  -R2.  Making  in  this  r  =  RI  we  have  the  total  pressure  of  the 
wire  envelope  on  the  tube  or 


and  solving  this  for  T  we  have  for  the  uniform  tension  of  wrapping 
to  produce  the  required  pressure  p\  on  the  exterior  of  the  tube 

2T>    O--/ 
... K\"P  i 

p  2       p  2  / 1  A^ 

/D  2  O   2\  1^^    **»      ~~   **«  ^iD/' 


If  the  jacket  is  assembled  without  shrinkage  on  the  wire  envelope, 
or  if  there  is  no  jacket,  2/1  becomes  the  total  pressure  Pip  required 
on  the  exterior  of  the  tube  to  compress  its  inner  surface  to  the 
elastic  limit  when  the  gun  is  in  the  state  of  rest. 

9.   Intensity  of  Pressure  at  any  Radius  r  of  the  Wire  Envelope 
Due  to  Wrapping,  System  at  Rest.  —  In  equation  (11)  pr  is  the 


ELASTIC  STRENGTH  OF  WIRE-WRAPPED  GUNS  11 

pressure  at  r  due  to  wrapping  from  r  to  any  other  radius  r'.  If 
r'  be  taken  equal  to  Rz,  pr  will  be  the  pressure  at  r  due  to  wrapping 
all  the  wire  from  r  out,  or  the  final  pressure  at  rest  at  r  due  to 
the  wrapping,  and  p't  —  Pt  will  be  the  pressure  on  the  exterior 
of  the  tube  produced  by  the  same  wrapping.  This  pressure  is 
the  p'tRt  of  equation  (14).  Substituting  for  p't  —  Pt  in  equation 
(11)  the  value  of  p'tRt  from  equation  (14)  we  have 

fl^-flo2  R2*  -  flo2  _  (RS  -  flo2)  r*pr 

ge 


2  R?  e  r*  -R<?   "    RS  (r2  -  #02) 

and  solving  for  pr 

_r^-Rl  R^-jRl 

Pr  -       2r2  g<  r2  -  R<? 

which  gives  the  pressure  pr  due  to  the  wrapping  at  any  radius  r 
of  the  wire  envelope  when  the  gun  is  in  the  state  of  rest.  If,  in 
equation  (17)  r  becomes  Rif  pr  becomes  p\  the  total  pressure  of 
the  wire  envelope  on  the  tube.  Making  these  substitutions  in 
equation  (17) 


the  same  as  given  in  equation  (15),  as  it  should  be. 

10.  Intensity  of  Tension  at  any  Radius  r  of  the  Wire  Envelope 
Due  to  Wrapping,  System  at  Rest.  —  Let  tT  be  the  tension  at  r 
due  to  the  wrapping  after  it  is  completed.  Then  calling  r  the 
mean  tension  of  the  wires  between  r  and  r', 

ii-  (i  P'rr'  —  prr 

p'rr'  -prr  =  -  T  (r'  -  r)     or         r>  _*     =  -r 

or  passing  to  the  limit,  when  T  becomes  —  tr, 

d  (prr) 


dr 


(18) 


Substituting  in  equation  (18)  the  value  of  pr  from  equation  (17) 
we  have 

/  H  r2  -  R°*  Tine,  ^  ~  flo2"| 

~tr  =  d|'~27~riQg'r'--/g,«l 
dr 


12  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

Performing  the  differentiation  indicated 

Rz2  -  .Ro2  [4  r*dr  -  (r2  -  fl02)  2  dr 


, 
'tr~  dr  10g< 


2  - 


flo2 1] 

-  #o2)J 


2r 
and  reducing 

r  *-2  J_   P.2  7?.2   _    P.2~| 

(19) 


r2  _L   P  2  P  2   P  1 

I         1^  J.VQ    |  J.l»2  -il'O 


Equation  (19)  gives  the  tension  due  to  the  wrapping  at  any  radius 
r  of  the  wire  envelope  when  the  gun  is  in  the  state  of  rest. 

11.   Tangential  and  Radial  Stresses  and  Strains  at  any  Radius 
r  of  the  Wire  Envelope  Due  to  Wrapping,  System  at  Rest.  —  The 

tangential  stress  at  any  radius  r  of  the  wire  envelope  due  to  the 
wrapping,  system  at  rest,  is 

Stw    =   tr  +  Pr/3  (20) 

and  substituting  in  this  the  values  of  tr  and  pr  from  equations 
(19)  and  (17) 


reducing 

r        r*  +  2flo2        RS-R<?-\ 
S^=T\I        -g^-   loge  r2  _  R,  J 

The  radial  stress  at  any  radius  r  of  the  wire  envelope  due  to  the 
wrapping,  system  at  rest,  is 

8^=    -pr-   tr/Z  (22) 


and  substituting  in  this  the  values  of  pr  and  tr  from  equations 
(17)  and  (19) 


T  -o  -     < 

~    ~  ~^    l°gt    22 


The  tangential  and  radial  strains  can  be  obtained    by  dividing 
Stu>  and  Spw,  respectively,  by  the  modulus  of  elasticity  E. 

12.  Stresses  in  Jacket,  Wire  Envelope,  and  Tube,  System  at 
Rest  and  in  Action.  —  Equations  (17),  (19),  (21),  and  (23)  give 
the  radial  pressure,  the  tangential  tension,  the  tangential  stress, 


ELASTIC  STRENGTH  OF  WIRE-WRAPPED  GUNS 


13 


and  the  radial  stress  due  to  the  wrapping,  at  any  radius  r  of  the 
wire  envelope  when  the  system  is  at  rest,  in  terms  of  the  uniform 
tension  of  wrapping  T  and  the  radii  of  the  tube  and  wire  envelope. 
The  shrinkage  of  the  jacket  also  produces  radial  pressure,  tan- 
gential tension,  and  tangential  and  radial  stresses  in  the  wire 
envelope  which  must  be  added  algebraically  to  those  produced 
by  the  wrapping  to  obtain  the  final  state  of  the  wire  envelope, 
system  at  rest.  The  stresses  in  the  jacket,  system  at  rest,  are 
due  only  to  the  shrinkage  pressure  P'2  and  those  in  the  tube, 
system  at  rest,  only  to  the  exterior  pressure  on  the  tube  Pip. 
The  radial  pressure,  tangential  tension,  and  tangential  and  radial 
stresses  in  both  jacket  and  tube,  system  at  rest,  can  be  calculated 
from  equations  (4),  (3),  (1),  and  (2),  respectively,  after  proper 
substitutions  therein. 

The  tangential  tension,  radial  pressure,  and  stresses  in  the 
tube,  wire  envelope,  and  jacket  having  been  calculated  for  the 
gun  in  the  state  of  rest,  the  tensions,  pressures,  and  stresses  in 
action  corresponding  to  any  pressure  P0  may  be  obtained  by  add- 
ing algebraically  to  those  in  the  state  of  rest  the  tensions,  pressures, 
and  stresses  computed  from  equations  (3),  (4),  (1),  and  (2), 
respectively,  by  considering  the  gun  as  a  single  cylinder  acted  on 
only  by  an  interior  pressure  P0. 


Fig.  2. 

EXAMPLE. 

13.  Figure  2  represents  a  section  through  the  powder  chamber 
of  a  12-inch  wire-wrapped  gun. 

The  material  and  physical  qualities  of  the  tube,  wire  envelope 
and  jacket  are  shown  in  the  following  table: 


14 


STRESSES  IN  GUNS   AND  GUN  CARRIAGES 


Part 

Material 

Elastic  limit 
in  tension 

e 

Elastic  limit 
in  compression 
p 

Modulus  of 
elasticity 
E 

Tube 

Forged  steel 

45,000 

60,000 

30,000,000 

Wire 

Steel  wire 

100,000 

100,000 

30,000,000 

Jacket 

Cast  steel 

30,000 

30,000 

30,000,000 

It  is  required  to  determine, 

(a)  the  maximum  powder  pressure  which  the  gun  may  with- 
stand without  the  tangential  stresses  in  any  part  thereof 
exceeding  its  elastic  limit; 

(6)  the  shrinkage  of  the  jacket; 

(c)  the  uniform  tension  of  wrapping  of  the  wire  envelope; 

(d)  the  stresses  at  the  interior  and  exterior  surfaces  of  each  part 

in  the  state  of  rest;  and 

(e)  the  stresses  at  the  interior  and  exterior  surfaces  of  each  part 

in  the  state  of  action  corresponding  to  the  maximum  per- 
missible powder  pressure  and  also  to  a  powder  pressure  of 
42,000  Ibs.  per  sq.  in. 

Maximum    Permissible    Powder    Pressure  —  Shrinkage    of 
Jacket  —  Uniform  Tension  of  Wrapping.  —  From  Fig.  2 

RQ  =  6;  #!  =  9;  #2  =  13.15;  R3  =  18. 
From  equation  (5)  the  maximum  permissible  powder  pressure  is 


3[(18)2  -  (6)2]  [60,000  +  45,000] 


=  66,316  Ibs.  per  sq.  in.  (24) 


4  (18)2  +  2  (6)2 

The  pressure  on  the  exterior  of  the  .tube  to  compress  its  interior 
surface  in  the  state  of  rest  to  the  elastic  limit  in  compression, 
60,000  lbs.  per  sq.  in.,  is,  from  equation  (6), 


60>000  =  16>667  lbs- 


,au 

(y) 


Part  of  this  pressure  is  due  to  the  shrinkage  of  the  jacket. 
To>  determine  this  shrinkage  it  is  first  necessary  to  find  the  in- 
tensity of  stress  at  the  interior  of  the  jacket  in  the  state  of  action 
due  to  the  powder  pressure  of  66,315  lbs.  per  sq.  in.,  and  to  sub- 


ELASTIC  STRENGTH  OF  WIRE-WRAPPED  GUNS  15 

tract  this  stress  from  the  elastic  limit  in  tension  of  the  material 
in  the  jacket.  The  difference  will  be  due  to  the  shrinkage  pres- 
sure and  from  it  the  shrinkage  pressure  can  be  determined.  Hav- 
ing the  shrinkage  pressure,  the  pressure  which  it  transmits  to  the 
exterior  of  the  tube  may  be  obtained.  Substituting  in  equation 
(1)  #3  =  18  for  Ri,  R2  =  13.15  for  r,  and  making  PI.  =  0  and 
Po  =  66,316,  the  tangential  stress  at  the  inner  surface  of  the 
jacket  due  to  P0  is 

„„    _  2  66,316  (6)2       4       (6)2  (18)2  66,316 

«J     «3  o     /1ON9  /C\<>.     •     O 


3  (18)2  -  (6)2  '  3  [(18)2  -  (6)2]  (13.15)2 
=  26,235  Ibs.  per  sq.  in.  tension.  (26) 

The  elastic  limit  in  tension  of  the  jacket  being  30,000  Ibs.  per 
sq.  in.,  30,000  -  26,235  or  3765  Ibs.  per  sq.  in.  will  be  the  stress 
due  to  the  shrinkage  pressure.  Substituting  in  equation  (1), 
R2  =  13.15  for  R0,  #3  =  18  for  Rlf  making  Pl  =  0,  P0  =  P'2,  and 
St  =  3765,  and  solving  for  P'2,  the  shrinkage  pressure, 


P'2  =  3?65  =  1039  lbs'  Per  Sq'  in'      (27) 


The  absolute  shrinkage  to  produce  this  pressure  is  from  equation 

(8) 

c  4  (13.15)3  [(18)2  -  (6)2]  1039 

02  = 


30,000,000  [(13.15)2  -  (6)2]  [(18)2  -  (13.15)2] 
=  .00439  inch.  (28) 

The  pressure  on  the  exterior  of  the  tube  due  to  the  shrinkage 
pressure  of  the  jacket  from  equation  (9),  derived  from  equation 
(4),  is 

2  =  729  lbs.  per  sq.  in.     (29) 


Subtracting  this  from  the  total  exterior  pressure  on  the  tube, 
equation  (25),  when  the  system  is  at  rest,  we  have  p'i  =  16,667  — 
729  =  15,938  lbs.  per  sq.  in.  as  the  part  of  the  pressure  on  the 
exterior  of  the  tube,  system  at  rest,  which  must  be  due  to  the 
wire  envelope.  Substituting  this  value  of  p'i  in  equation  (16) 


9  fQ">2  1 

T  =  -  =  51>566  lbs-  Per  «l-  in-    *(30> 


(9)2  -  (6)2 


*  The  Naperian  logarithm  of  a  number  is  equal  to  the  common  logarithm 
of  the  number  divided  by  .4343. 


16  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

which  is  the  uniform  tension  under  which  the  wire  must  be 
wrapped  on  the  tube. 

STRESSES  AT  BEST. 
(See  figures  3  and  4.) 

14.  Inner  Surface  of  Tube.  —  The  tangential  stress,  see 
equation  (25),  is  one  of  compression  equal  to  60,000  Ibs.  per  sq. 
in.  The  radial  stress  obtained  from  equation  (2)  by  making 
P0  =  0,  Pi  =  16,667,  r  =  R0  =  6,  and  #1  =  9,  is 

„       2  -16,667  (9)2      4  -  (9)2  16,667 


p      3    (9)2  -  (6)2        3     (9)2  -  (6)2 
=  +20,000  Ibs.  per  sq  .in.  (31) 

and  since  the  result  is  positive  the  stress  is  one  of  tension. 

Outer  Surface  of  Tube.  —  The  tangential  stress,  obtained  from 
equation  (1)  by  making  P0  =  0,  Pi  =  16,667,  R0  =  6,  and  r  =  RI  = 
9,  is 
0       2  -  16,667  (9)2    ,  4  -(6)2  16,667  0 


t      3    (9)2  -  (6)2        3     (9)2  -   6)'         ~ 

(32) 

and  since  the  result  is  negative  the  stress  is  one  of  compression. 
The  radial  stress,  obtained  by  making  the  same  substitutions  in 
equation  (2),  is 

2  -  16,667  (9)2      4  -(6)2  16,667 
S'  =  3    (9)2-(6)2     -  3     (9)2-(6)2     -  ~2223  lbs'  per  sq'  m' 

(33) 
and  the  stress  is  one  of  compression. 

Inner  Surface  of  Wire  Envelope,  —  The  tangential  stress  due 
to  the  layers  of  wire  only,  obtained  from  equation  (21)  by  making 
Ro  =  6,  r  =  RI  =  9,  R2  =  13.15,  and  T  =  51,566,  see  equation 
(30),  is 

(9)2  +  2(6)2        (13.15)2  -  (6)2 


3(9)2  ,      (9)2_(6)2 

=  +15,439  lbs.  per  sq.  in.  (34) 

and  the  stress  is  one  of  tension. 

This  stress  is  decreased  by  that  due  to  the  shrinkage  pressure 
of  the  jacket  which  is  obtained  from  equation  (1)  by  making 

Ro  =  6,R1  =  R2  =  13.15,  r  =  R1  =  9,P1  =  P'2  =  1039,  and  P0  =  0. 


ELASTIC  STRENGTH  OF  WIRE-WRAPPED  GUNS  17 

Whence 

„„        2  -(1039)  (13.15)2       4  -(6)2(13.15)2(1039)         1 

«J       tW  O  /•«  O    1  C\'>  //?\9  I      O 


3     (13.15)2  -  (6)2     '  3        (13.15)2  -  (6)2  (9)2 

=  — 1653  Ibs.  per  sq.  in.  compression.  (35) 

The  final  tangential  stress  at  rest  is,  therefore, 

15,439  -  1653  =  13,786  Ibs.  per  sq.  in.  tension. 

The  radial  stress  due  to  the  layers  of  wire  only,  obtained  from 
equation  (23)  by  making  the  same  substitutions  as  in  equation 
(21)  for  the  tangential  stress,  is 

51,566  [        (9)2-2(6)2        (13.15)2  -  (6)2] 
3      {  (9)2  (9)2  -  (6)2    ] 

=  — 19,314  Ibs.  per  sq.  in.  compression.  (36) 

To  this  must  be  added  algebraically  the  radial  stress  due  to 
the  shrinkage  of  the  jacket  which  is  obtained  from  equation  (2) 
by  making  the  appropriate  substitutions  therein. 

Whence 
„„      _  2  -(1039)  (13.15)2      4  -(6)2  (13.15)2  (1039):  ^    1 

O       -mn  o 


3     (13.15)2  -  (6)2        3        (13.15)2  -  (6)2  (9)2 

=  —  97  Ibs.  per  sq.  in.  compression.  (37) 

The  final  radial  stress  at  rest  is,  therefore, 

—  19,314  —  97  =  —  I9,4ll  Ibs.  per  sq.  in.  compression. 

Outer  Surface  of  Wire  Envelope.  —  Before  the  assembling  of 
the  jacket  the  tangential  stress  in  51,566  Ibs.  per  sq.  in.,  the 
tension  of  wrapping.  This  is  decreased  by  the  compression  due 
to  the  shrinkage  of  the  jacket  which  is  obtained  from  equation 
(1)  by  making  the  proper  substitutions  therein. 
Whence 

„„     _2  -  (1039)  (13.15)2      4  -(6)2(13.15)2(1039) 
J  tw  ~  3    (13.15)2  -  (6)2     *"  3        (13.15)2  -  (6)2 

x  ,..q  -ic\2  =  — 1240  Ibs.  per  sq.  in.  compression.         (38) 

The  final  tangential  stress  at  rest  is,  therefore, 

51,566  -  1240  =  50,326  Ibs.  per  sq.  in.  tension. 


18  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

Before  assembling  the  jacket  the  radial  stress  is  minus  one- 
third  the  tension  of  wrapping  =  —51,566X3  =  — 17,188  Ibs.* 
per  sq.  in.  compression.  The  radial  stress  due  to  the  shrinkage 
of  the  jacket,  obtained  from  equation  (2)  by  making  the  proper 
substitutions  therein,  is 

„„      _  2  -(1039)  (13.15)2  _  4  -(6)2(13.15)2(1039)  >         1 
'  ""      3    (13.15)2  -  (6)2        3        (13.15)2  -  (6)2  (13.15)2 

=  —510  Ibs.  per  sq.  in.  compression.  (39) 

The  final  radial  stress  at  rest  is,  therefore, 

-  17,188  —  510  =  —17,698  Ibs.  per  sq.  in.  compression. 

Inner  Surface  of  Jacket.  —  All  stresses  in  the  jacket,  system 
at  rest,  are  due  to  the  shrinkage.  The  tangential  stress  at  its 
inner  surface,  see  equation  (27),  is  3765  Ibs.  per  sq.  in.  tension. 
The  radial  stress,  obtained  from  equation  (2)  by  making  the  proper 
substitution  therein,  is 

2   (1039)  (13.15)2       4  (13.15)2  (18)2  (1039) 


£1  t-t       ^  j.v/tJt/y    ^  j.t-/»  j.t/y  T:     v^  -*-*-'• -*•*-'/      *"^/      V     "t-Jt/  y  _L 

w'  =  3  (18)2  -  (13.15)2  ~  3      (18)2  -  (13.15)2         (13.15)2 
=  —2178  Ibs.  per  sq.  in.  compression.  (40) 

Outer  Surface  of  Jacket.  —  The  tangential  stress,  obtained 
from  equation  (1)  by  making  the  proper  substitutions  therein, 
is 

„    _  2  (1039)  (13.15)2       4  (13.15)2  (18)2  (1039)         1 
<y      3  (18)2  -  (13.15)2  +  3      (18)2  -  (13.15)2         (18)2 
=  +2378  Ibs.  per  sq.  in.  tension.  (41) 

The  radial  stress,  obtained  from  equation  (2)  by  making  the 
proper  substitutions  therein,  is 

2   (1039)  (13.15)2      4  (13.15)2  (18)2  (1039) 


«  *-<         V,  -*-  VftJl/y     ^  -LtJ*  JLU  J  ~       \^  J.  ?_J  •  J- C/ y        y.LWy        \J.\JU*S  J  J. 

pj  =  3  (18)2  -  (13.15)2  ~  3      (18)2  -  (13.15)2          (I8)2 
=  —793  Ibs.  per  sq.  in.  compression.  (42) 


*  This  result  can  also  be  obtained  from  equation  (23)  by  proper  substitu- 
tions therein. 


ELASTIC  STRENGTH  OF  WIRE-WRAPPED  GUNS  19 

• 

STRESSES  IN  ACTION. 
(See  figures  3  and  4.) 

15.   Maximum  Permissible  Powder  Pressure  P0  =  66,315.  — 

Making  in  equations  (1)  and  (2)  P0  =  66,315,  Pi  =  0,  R0  =  6, 
and  RI  =  R3  =  18,  they  become,  respectively, 


66,315  (6)2 


(18)2  -  (6)2 


I  +  4fl|)![  =  5526+  [6.55400].       (43) 


and 


66,315  (6)2  [2  _  4  (18)2]  _          _  [6.55400]* 
"  (18)2  -  (6)2[3      3     r2   J  =  r2 


By  substituting  for  r  in  equations  (43)  and  (44)  its  values  corre- 
sponding to  the  surfaces  of  the  various  parts  of  the  gun,  the 
stresses  due  to  the  action  of  P0  on  the  gun  considered  as  a  single 
cylinder  may  be  determined,  and  the  algebraic  sums  of  these 
stresses  and  those  previously  determined  for  the  state  of  rest  are 
the  stresses  in  action. 

Inner  Surface  of  Tube,  r  =  R0  =  6.  —  The  tangential  stress 
system  in  action,  see  equation  (24),  is  45,000  Ibs.  per  sq.  in.  ten- 
sion. The  radial  stress  is 


+20,000  +  5526  -     '2       =  +20,000-93,946  =  -73,9461bs.f 

per  sq.  in.  compression.         (45) 

Outer  Surface  of  Tube,     r  =  RI  =9.  —  The  tangential  stress 
is 


-37,777  +  5526  +  ±     ^      =  -37,777  +  49,735  =  +11,958  Ibs. 

per  sq.  in.  tension.        (46) 
The  radial  stress  is 

-2223  +  5526  -  [6'^f°]  =  -2223  -  38,683  =  -40,906  Ibs. 

per  sq.  in.  compression.        (47) 

*  The  figures  in  brackets  are  logarithms  of  the  numbers, 
t  This  exceeds  the  elastic  limit  for  compression. 


20  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

i 

Inner   Surface  of  Wire   Envelope,    r  =  RI  =  9.  —  The  tan- 
gential stress  is 


+13,786  +  5526  +     '(2      =  +13,786  +  49,735  =  +63,521  Ibs. 

per  sq.  in.  tension.         (48) 
The  radial  stress  is 

-19,411  +  5526  -  [6'^t°°]  =  -19,411  -  38,683  =  -58,096  Ibs. 
W 

per  sq.  in.  compression.         (49) 

Outer    Surface    of    Wire    Envelope,    r  =  R2  =  13.15.  —  The 

tangential  stress  is 


+50,326  +  5526  +      g2  =  +50,326+26,234  =  +76,560  Ibs. 

per  sq.  in.  tension.         (50) 
The  radial  stress  is 

-17,698  +  5526  -  ^fg^j   =  -17,698  -  15,182  =  -32,880  Ibs. 

per  sq.  in.  compression.         (51) 

Inner  Surface  of  Jacket,    r  =  R2  =  13.15.  —  The  tangential 
stress  is,  see  equations  (26)  and  (27), 

+3765  +  26,234  =  +29,999  Ibs.  per  sq.  in.  tension.       (51|) 
The  radial  stress  is 


-2178  +  5526  -  .  =  -2178  -  15,182  =  -17,360  Ibs. 

\Lo.Lo) 

per  sq.  in.  compression.         (52) 

Outer    Surface    of    Jacket,    r  =  Rs  =  18.  —  The    tangential 
stress  is 


+£378  +  5526  +     '  =  +2378  +  16,578  =  +18,956  Ibs. 

per  sq.  in.  tension.         (53) 
The  radial  stress  is 

-793  +  5526  -    ^f-fg^  =  -793  -  5526  =  -6319  Ibs. 

per  sq.  in.  compression.        (54) 

The  tangential  stresses,  system  at  rest,  and  in  action  when 
P0  =  66,315  Ibs.  per  sq.  in.  are  shown  graphically  in  Fig.  3,  and 
the  radial  stresses  in  Fig.  4. 


ENGINEERING 


ELASTIC  STRENGTH  OF  WIRE-WRAPPED  GUNS  21 


105000 


.45000 


j* 

I- 


60000 


Bore. 
—  6.0- 


Tantrenlial  Stresses 
*  Pa-663l6 


Tube. 


4—  3.0- 


W786 


3765 


37777 


Wire  Envelope. 


4.15 


76578 


50326 


30000 


at  r*> 


Jacket. 
-  4.85  — 


18956 
16578 


2378 


Fig.  3. 
Radial  Stresses. 


Fig.  4. 


22  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

STRESSES  IN  ACTION 

(See  figures  5  and  6.) 

16.  Po  =  42000  Ibs.  per  sq.  in.  —  By  reference  to  equations  (1) 
and  (2),  or  to  equations  (43)  and  (44),  it  will  be  seen  that  the  stresses 
due  to  the  action  of  an  interior  pressure  on  a  gun  considered  as  a 
single  cylinder  are  directly  proportional  to  the  intensity  of  the 
interior  pressure.  The  stresses"  at  the  inner  and  outer  surfaces 
of  the  various  parts  of  this  gun,  system  in  action,  under  an  in- 
terior powder  pressure  of  42,000  Ibs.  per  sq.  in.  will,  therefore, 
be  the  stresses  at  rest  plus  42,000X66,316  of  the  increases  in 
stress  in  the  various  parts  due  to  the  action  on  the  gun,  considered 
as  a  single  cylinder,  of  the  maximum  permissible  powder  pressure 
of  66,316  Ibs.  per  sq.  in.  These  increases  in  stress  have  already 
been  calculated  and  are  given  in  equations  (45)  to  (54),  inclusive, 
excepting  the  increase  in  tangential  stress  at  the  bore  of  the  tube 
which  had  been  previously  determined  to  be  105,000  Ibs.  per  sq.  in. 

Inner  Surface  of  Tube.  —  The  tangential  stress  is 

4.9  000 
-60,000  +l§^i  x  105,000  =  -60,000  +  66,501  =  +6501  Ibs. 

OO,olO 

per  sq.  in.  tension          (55) 
The  radial  stress  is 


+20,000  +  x(  -93,946)  =  +20,000  -59,500=  -39,500  Ibs. 

bb,olo 

per  sq.  in.  compression.         (56) 
Outer  Surface  of  Tube.  —  The  tangential  stress  is 

49  00ft 

-37,777  +  li^  X  49,735  =  -37,777  +  31,499  =  -6278  Ibs. 
bb,olb 

per  sq.  in.  compression.          (57) 
The  radial  stress  is 


-2223  +        T5  x  (-38,683)  =  -2223  -  24,500  =  -26,723  Ibs. 

bb,olb 

per  sq.  in.  compression.         (58) 
Inner  Surface  of  Wire  Envelope.  —  The  tangential  stress  is 


+13,786  +  x  49,735  =  +13,786  +  31,499  =  +45,285  Ibs. 

per  sq.  in.  tension.        (59) 


ELASTIC  STRENGTH  OF  WIRE-WRAPPED  GUNS  23 

The  radial  stress  is 


-19,411  +    =££  x  (-38,683)  =  -19,411-24,500  =  -43,911  Ibs. 

bb,olb 

per  sq.  in.  compression.         (60) 
Outer  Surface  of  Wire  Envelope.  —  The  tangential  stress  is 


+50,326  +  x  26,234  =  +  50,326  +  16,615  =  +66,941  Ibs. 

bb,olb 

per  sq.  in.  tension.         (61) 
The  radial  stress  is 


-17,698  +  ~        X  (-15,182)  =  -17,698-  9616  =  -27,314  Ibs. 

bb,olb 

per  sq.  in.  compression.         (62) 

Inner  Surface  of  Jacket.  —  The  tangential  stress  is 
49  000 

+3765  +  ~^  x  26,234  =  +3765  +  16,615  =  +20,380  Ibs. 
bb, 


per  sq.  in.  tension.         (63) 
The  radial  stress  is 

A.9  000 

-2178  +li^  x  (-15,182)  =  -2178  -  9616  =  -11,794  Ibs. 


per  sq.  in.  compression.         (64) 
Outer  Surface  of  Jacket.  —  The  tangential  stress  is 


x  16,578  =  +2378  +  10,502  =  +12,880  Ibs. 

bb,olb 

per  sq.  in.  tension.         (65) 
The  radial  stress  is 


-793  +  x  (-5526)  =  -793  -  3500  =  -4293  Ibs. 

bb,olb 

per  sq.  in.  compression.         (66) 

Figs.  5  and  6  show  graphically  the  tangential  and  radial 
stresses,  respectively,  system  in  action,  when  P0  =  42,000  Ibs. 
per  sq.  in. 


24 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


Tangential  Stressed. 
P0  -  42000 


Fig.  5. 


Radial  Stresses. 
P0=  42000 


.at  rest. 


3500 
4293 


24500.  ^i 


at  rest. 


JS500 


55500 


26723 


27314 


•6.0- 


Tube. 
(-3.0- 


Wire  Envelope. 
-4.15 — 


Jacket. 
-4.85— 


Fig.  6. 


ELASTIC  STRENGTH  OF  WIRE-WRAPPED  GUNS 


25 


17.  Problem.  —  Fig.  7  shows  a  longitudinal  section  through 
a  part  of  the  6-inch  wire-wrapped  gun  model  1908  just  in  front 
of  the  forcing  cone. 


Fig.  7. 


Part 

Prescribed 
elastic  limit 
p  =  fl 

Tube 
Wire  envelope 
Jacket 

50,000 
140,000 
50,000 

The  jacket  is  assembled  with  an  absolute  shrinkage  of  .007  inch. 
The  wire  is  wrapped  under  a  constant  tension  of  50,000  Ibs.  per 
sq.  in. 

Find: 

(a)  the  maximum  powder  pressure  which  the  gun  can  withstand 
without  the  tangential  stress  on  any  part  exceeding  the 
elastic  limit  in  tension  or  compression; 

(6)  the  tangential  and  radial  stresses  at  the  inner  and  outer  sur- 
faces of  each  part,  system  at  rest; 

the  tangential  and  radial  stresses  at  the  inner  and  outer  sur- 
faces of  each  part,  system  in  action,  when  subjected  to  the 
powder  pressure  determined  under  (a). 


(c) 


CASE  n.  —  WIRE  WRAPPED  UNDER  SUCH  VARYING  TENSION 
THAT  WHEN  THE  GUN  IS  FIRED  WITH  THE  PRESCRIBED 
MAXIMUM  POWDER  PRESSURE  ALL  LAYERS  OF  WIRE  WILL 
BE  SUBJECTED  TO  THE  SAME  TANGENTIAL  STRESS. 

18.   General  Method  of  Design.  —  Let  it  be  required  to  con- 
struct a  wire-wrapped  gun  as  shown  in  Fig.  1  or  2  in  such  manner 


26  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

as  to  give  it  the  maximum  elastic  strength  permitted  by  the  qual- 
ity of  the  metal  in  the  tube  and  the  total  thickness  of  wall  P3  —  R0, 
and  with  the  condition  that  when  the  gun  is  fired  with  the  pre- 
scribed maximum  powder  pressure  all  layers  of  wire  shall  be  sub- 
jected to  the  same  tangential  stress. 

First  determine  from  equation  (5)  the  maximum  interior 
pressure  which  the  gun  can  support  in  action  on  the  assumption 
that  in  the  state  of  rest  the  inner  surface  of  the  tube  is  compressed 
to  its  elastic  limit.  From  equation  (6)  determine  the  pressure 
PIP  on  the  exterior  of  the  tube  which  will  compress  its  inner  sur- 
face in  the  state  of  rest  to  its  elastic  limit.  With  the  values  of 
Po  and  PIP  from  equations  (5)  and  (6)  determine  from  equation 
(48),  page  215,  Lissak's  Ordnance  and  Gunnery,  the  correspond- 
ing value  of  the  pressure  on  the  exterior  of  the  tube,  system  in 
action.  The  last  equation,  after  replacing  the  radius  ratios  by 
their  values  in  terms  of  the  radii  and  substituting  R3  for  R2  since 
there  are  three  parts  to  the  gun  instead  of  two,  becomes 

.  Po2  (P32  -  RS)  PQ 

Fl  "  **  4     RS  (P32  -  Po2) 

The  next  steps  require  the  use  of  formulas  (e)  to  (0),  inclusive, 
referred  to  in  article  6  which  will  now  be  deduced.  Let  Stwa  be 
the  uniform  tangential  stress  in  the  wire,  system  in  action;  Sp^ra 
the  radial  stress  at  any  radius  r  of  the  wire  envelope,  system  in 
action;  tra  the  tension  at  any  radius  r  of  the  wire  envelope,  system 
in  action;  pra  the  pressure  at  any  radius  r  of  the  wire  envelope, 
system  in  action;  and  Tr  the  tension  of  wrapping. 

19.  Radial  Pressure  at  any  Radius  r  of  the  Wire  Envelope, 
System  in  Action.  Constant  Tangential  Stress  in  the  Wire 
Envelope,  System  in  Action.  —  Substituting  for  tra  in  the  ex- 

pression 

=  ~tradr 


its  value,  tra  —  Stwa  —  pro/3  obtained  from  the  first  of  equations 
(4),  page  192,  Lissak's  Ordnance  and  Gunnery,  by  considering 
q  =  0,  we  have 

Pradr  +  rdpra  =  -  (Stmt  -  pro/3)  dr  (68) 

Whence 

0       *%"  —  TO  =  -dr/r  (69) 

Stwa  +  2  Pro/3 


ELASTIC  STRENGTH  OF  WIRE-WRAPPED  GUNS  27 

Integrating 

|  log,  (Stwl  +  f  pra}  =  log,  C/r  (70) 

Since  the  wire  envelope  is  covered  by  a  jacket  there  will,  when 
the  system  is  in  action,  be  a  pressure  P2  on  the  exterior  of  the 
wire  envelope,  which  pressure,  if  the  full  elastic  strength  of  the 
jacket  is  to  be  utilized,  must  be  such  as  to  extend  its  inner  surface, 
system  in  action,  to  its  elastic  limit.  Therefore,  when  r  =  R2, 
Pra  =  Pa;  and  the  constant  of  integration  C  is 

(£«.  +  !  P.)*  ft 

Substituting  this  value  of  C  in  equation  (70) 

I  log,  [Stwa  +  f  pra]  =  log,  [(Stua  +  f  P«)H  R2/r] 
or 

(Stwa  +  f  pm)*  =  (Stwa  +  f  P,)*  R2/r 

and  solving  for  pra 

Pra  =  I  [(Stwa  +  f  P2)  (&/r)*  -  Stwa]  (71) 

which  gives  the  radial  pressure  at  any  radius  r  of  the  wire  envelope, 
system  in  action.  Making  in  this  r  =  RI,  the  pressure  on  the 
exterior  of  the  tube,  system  in  action,  is 

Pi  =  I  [(Stwa  +  f  P2)  (ft/flO*  -  Stwa]  (72) 


and  solving  for  Stwa  we  obtain  the  constant  tangential  stress  in 
the  wire  envelope,  system  in  action,  required  to  produce  a  given 
pressure  PI  on  the  exterior  of  the  tube,  system  in  action. 
Whence 


_ 


2  Pi  -  P2 


3      £2fliK  -  1 


20.  Tangential  Tension  and  Radial  Stress  at  any  Radius  r  of 
the  Wire  Envelope,  System  in  Action.  —  The  tangential  tension 
in  at  any  radius  r  of  the  wire  envelope,  system  in  action,  is  tra  = 
Stwa  —  Pro/%  and  substituting  in  this  the  value  of  pra  from 
equation  (71) 

tra    =   I  Stoa   ~   %  (Stwa  +  f  P2)   (fl2/>)*  (74) 

The  tangential  stress  Stwa  throughout  the  wire  envelope, 
system  in  action,  is  constant  by  hypothesis.  The  radial  stress 
at  any  radius  r  of  the  wire  envelope,  system  in  action,  is,  from  the 


28  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

second  of  equations  (4),  page  192,  Lissak's  Ordnance  and  Gun- 
nery, when  q  is  taken  as  zero, 

Spwra    =    —  (Pra  +  tra/3) 

and  substituting  for  pra  and  tn  their  values  from  equations  (71) 
and  (74),  respectively, 

S^ra   -    -  f  (Stwa  +  f  Po)   (fi,/r)*  +  Stwa  (75) 

The  tangential  and  radial  strains  can  be  obtained  by  dividing 
the  corresponding  stresses  by  30,000,000,  the  modulus  of  elas- 
ticity E  for  steel. 

21.  Variable  Tension  of  Wrapping  at  any  Radius  r  of  Wire 
Envelope.  —  Equation  (74)  gives  the  tangential  tension  tra  at 
any  radius  r  of  the  wire  envelope,  system  in  action.  The  increase 
in  tension  at  the  radius  r,  system  in  action,  over  the  tension  of 
wrapping  Tr  is  due  to  the  interior  pressure  P0  and  the  pressure 
Pra,  system  in  action,  neither  of  which  was  acting  when  the  layer 
at  the  radius  r  was  applied.  Considering  the  part  of  the  gun 
between  the  interior  radius  R0  and  the  radius  r  of  the  wire  envelope 
as  a  single  cylinder,  the  increase  in  tension  at  the  radius  r  over 
the  tension  of  wrapping  due  to  the  pressures  P0  and  pra  is  obtained 
from  equation  (3)  by  making  PI  =  pn  and  R\  =  r. 

Whence 


2  - 


Subtracting  this  increase  in  tension  from  the  tension  at  the 
radius  r  of  the  wire  envelope,  system  in  action,  there  results  the 
tension  of  wrapping  at  any  radius  r 


»  E>  2 

f*   —  /to 

Substituting  in  this  the  values  of  tra  and  pra  from  equations  (74) 
and  (71),  respectively, 

Tr  =  |  Stwa  -  \  (Stwa  +  f  Po)  (Rz/r)K 
o2  -  I  [(Stwa  +  I  P2)  (R*/r}K  -  Stwa]  (r2  +  #02) 


r2  - 
and  reducing 


P2)  (R*/r}K  (r2  +  2  fl02)  -  #02  (3  £ 


„  t»a  2       *  0     -     0         toa0 

Tr  =  -  -  (76) 


ELASTIC  STRENGTH  OF  WIRE-WRAPPED  GUNS  29 

The  tension  of  wrapping  at  any  radius  r  may  be  obtained  from 
equation  (76)  by  substituting  therein  the  particular  value  of  r. 

22.  Stresses  in  the  Jacket,  the  Wire  Envelope,  and  the  Tube, 
System  in  Action  and  at  Rest.  —  The  radial  pressure,  constant 
tangential  stress,  tangential  tension,  and  radial  stress  in  the  wire 
envelope,  system  in  action,  may  be  obtained  by  proper  substitu- 
tions in  equations  (71),  (73),  (74),  and  (75),  respectively.    The 
stresses  in  the  jacket,  system  in  action,  are  due  only  to  the  pres- 
sure P2  on  its  inner  surface;   and  those  in  the  tube,  system  in 
action,  are  due  to  the  pressure  P0  on  its  inner  surface  and  the 
pressure  Pi  on  its  exterior.     The  tangential  tension,  radial  pres- 
sure, and  the  tangential  and  radial  stresses  in  both  jacket  and  tube, 
system  in  action,  may  be  calculated  from  equations  (3),  (4),  (l)t 
and   (2),  respectively,  after  proper  substitutions  therein.     The 
tangential  tensions,  radial  pressures,  and  stresses  actually  exist- 
ing in  the  various  parts  of  the  gun,  system  in  action,  correspond- 
ing to  an  interior  powder  pressure  P0  having  been  calculated, 
those  at  rest  may  be  obtained  by  subtracting  algebraically  from 
the  tensions,  pressures,  and  stresses  in  action  those  computed  from 
equations  (3),  (4),  (1),  and  (2),  respectively,  by  considering  the 
gun  as  a  single  cylinder  acted  on  only  by  an  interior  pressure  P0. 

EXAMPLE. 

23.  Let  it  be  required  to  determine  for  the  wire- wrapped  gun 
shown  in  Fig.  2 

(a)  the  maximum  powder  pressure  which  it  can  withstand  with- 
out the  tangential  stress  in  any  part  exceeding  its  elastic 
limit; 

(6)  the  shrinkage  of  the  jacket; 

(c)  the  varying  tension  of  wrapping  of  the  wire  envelope  so  that 

each  layer  shall  be  subjected  to  the  same  tangential  stress, 
system  in  action,  under  the  maximum  powder  pressure 
determined  under  (a); 

(d)  the  stresses  at  the  interior  and  exterior  surfaces  of  each  part, 

system  in  action,  under  the  maximum  powder  pressure; 

(e)  the  stresses  at  the  interior  and  exterior  surfaces  of  each  part, 

system  at  rest;  and 

(/)  the  stresses  at  the  interior  and  exterior  surfaces  of  each  part, 
system  in  action,  under  a  powder  pressure  of  42000,  Ibs. 
per  sq.  in. 


30  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

Maximum  Permissible  Powder  Pressure.  —  Pressure  on  the 
Exterior  of  the  Tube  and  on  the  Interior  of  the  Jacket,  System 
in  Action.  —  Shrinkage  of  the  Jacket.  —  The  maximum  permis- 
sible interior  pressure  in  this  case  is  the  same  as  when  the  wire 
was  wrapped  under  constant  tension  and  is 

P0  =  66,316  Ibs.  per  sq.  in.,  from  equation  (24). 

The  value  of  PIP  from  equation  (25)  is  16,667  Ibs.  per  sq.  in. 
Substituting  this  value  of  Pip  and  the  value  of  P0  =  66,316  in 
equation  (67),  the  value  of  the  exterior  pressure  on  the  tube, 
system  in  action,  is 


"  16-667  +  •  41-5351bs- 


The  elastic  limit  of  the  jacket  being  30,000  Ibs.  per  sq.  in.  the 
pressure  P2  in  action  between  the  jacket  and  the  wire  envelope 
which  will  extend  the  inner  surface  of  the  jacket  to  its  elastic 
limit  is  from  equation  (1),  after  making  the  proper  substitutions 
therein  and  solving  for  P2, 

3  [(18)2  -  (13.15)2]  30,000 
Pa=        4  (18)'  +  2(13.15)'        =82811bs*Perscl-in-     (?8) 

From  equation  (26)  the  tangential  stress  at  the  inner  surface  of 
the  jacket  due  to  the  action  of  the  powder  pressure  of  66,316  Ibs. 
per  sq.  in.  was  found  to  be  26,235  Ibs.  per  sq.  in.,  leaving  a  stress 
of  3765  Ibs.  per  sq.  in.  to  be  produced  by  the  shrinkage  of  the  jacket 
on  the  wire  envelope.  The  shrinkage  pressure  to  produce  this 
stress  was,  from  equation  (27),  found  to  be  1039  Ibs.  per  sq.  in., 
and  the  corresponding  absolute  shrinkage  from  equation  (28) 
was  .00439  inch. 

24.  Constant  Tangential  Stress  in  Wire  Envelope,  System  in 
Action.  —  Variable  Tension  of  Wrapping.  —  The  constant  tan- 
gential stress  in  the  layers  of  the  wire  envelope,  system  in  action, 
may  now  be  obtained  from  equation  (73)  by  substituting  therein 
the  values  of  PI  and  P2  from  equations  (77)  and  (78),  respec- 
tively; and  then  the  variable  tension  of  wrapping  from  equation 
(76)  by  substituting  therein  the  values  obtained  for  Stwa,  Pa,  Po, 
and  the  value  of  r  for  each  layer  of  the  wire  envelope. 


ELASTIC  STRENGTH  OF  WIRE-WRAPPED  GUNS  31 

Solving  equation  (73)  for  Stwa  after  substituting  for  PI  and  P2 
the  values  41,535  and  8281  from  equations  (77)  and  (78),  respec- 
tively, 

lJ41,535-828l(1^f( 

{15%    V  -  =  +71,578  Ibs.  per  sq.  in. 

-)    —  1  tension.         (79) 


which  is  the  constant  tangential  stress  in  all  the  layers  of  the 
wire  envelope,  system  in  action. 

Substituting   in  equation  (76)  the  values  of  St*,,  PQ,  P2,  RO, 
and  R2,  and  reducing 

(101  C\% 
==2)    (r2  +  72)  -  12,541,004 

•  ' 


By  substituting  in  equation  (80)  the  value  of  r  for  any  layer  of 
wire,  the  tension  of  wrapping  for  that  layer  may  be  obtained. 
The  wire  used  on  this  gun  was  of  square  cross  section  .1  inch  on 
a  side.  For  the  first  layer  of  wire  the  value  of  r,  taken  at  the 
center  of  the  thickness  of  the  wire,  is  9.05  and  the  corresponding 
value  of  Tr  =  58,425  Ibs.  per  sq.  in. 

For  the  middle  layer  of  wire,  the  value  of  r  taken  as  11.1  gives 
Tr  =  49,420  Ibs.  per  sq.  in. 

For  the  outside  layer,  the  value  of  r  taken  as  13.10  gives  Tr  = 
46,375  Ibs.  per  sq.  in. 

25.  Stresses  in  the  Jacket  and  the  Tube,  System  in  Action 
Po  =  66,316  Ibs.  per  sq.  in.,  and  at  Rest.  -  (See  Figs.  8  and  9.)  - 
Since  the  shrinkage  and  shrinkage  pressure  between  the  jacket 
and  wire  envelope,  and  the  pressure  on  the  exterior  of  the  tube, 
system  at  rest,  are  the  same  as  when  the  wire  envelope  was 
wrapped  under  a  constant  tension  of  51,566  Ibs.  per  sq.  in.,  the 
stresses  in  the  tube  and  jacket  both  in  action  and  at  rest  will  be 
the  same  as  determined  for  that  case. 


32  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

STRESSES  IN  THE  WIRE  ENVELOPE,  SYSTEM  IN  ACTION 
Po  =  66,316  Ibs.  per  sq.  In. 

(See  figures  8  and  9.) 

26.  The  tangential  stress  in  the  wire  envelope,  system  in 
action,  has  a  constant  value  of  71,578  Ibs.  per  sq.  in.     The  radial 
stress  Spwra  in  the  wire  envelope  may  be  obtained  by  substitution 
in  equation  (75)  of  the  proper  values  of  Stwa,  Pz,  #2,  and  r. 

Outer  Surface  of  the  Wire  Envelope,     r  =  R2  =  13.15,  Stwa  = 
71,578,  P2  =  8281. 
The  radial  stress  is 

-|  (71,578  +  |828l)  +  71,578  =  -31,220  Ibs.  per  sq.  in. 

compression.         (81) 

Inner    Surface   of   the    Wire    Envelope,     r  =  RI  =  9.  —  The 

radial  stress  is 

-4(71,578  +  \  828l)  (^^T+  71,578  =  -60,792  lb.persq.in. 
o\  o        /  \   y    / 

compression.         (82) 

STRESS  IN  THE  WIRE  ENVELOPE,   SYSTEM  AT  REST. 

(See  figures  8  and  9.) 

27.  The  tangential  and  radial  stresses  at  any  radius  r  of  the 
wire  envelope  due  to  the  action  of  P0  on  the  gun  considered  as  a 
single  cylinder  may  be  obtained  from  equations  (43)  and  (44), 
respectively,  by  the  substitution  therein  of  the  proper  values  of 
r.    Subtracting  the  results  thus  obtained  from  the  corresponding 
stresses  in  action,  we  have 

Outer  Surface  of  the  Wire  Envelope,    r  =  R2  =  13.15.  —  The 
tangential  stress  is 


+71,578  -     5526  +  =  +45,344  Ibs.  per  sq.  in. 

'  (-Lo.J.O_;      ) 

tension.         (83) 
The  radial  stress  is 


-31,220  -     5526  -  ^3  =  -16,038  Ibs.  per  sq.  in. 

compression.         (84) 


ELASTIC  STRENGTH  OF  WIRE-WRAPPED  GUNS 


33 


Inner    Surface   of  the    Wire   Envelope,    r  =  RI  =  9.  —  The 

tangential  stress  is 


+71,578-  5526  +     ' 
I 

The  radial  stress  is 


=  +21,843  Ibs.  per  sq.  in. 

tension.        (85) 


-60,792- 


[6.55400]] 
(9)2     J 


Ibs.  per  sq.  in. 
compression.        (86) 


Fig.  8. 


The  tangential  stresses,  system  in  action  when  P0  =  66,316  Ibs. 
per  sq.  in.,  and  system  at  rest,  are  shown  graphically  in  Fig.  8; 
and  the  radial  stresses  in  Fig.  9. 


34  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

Radial  Stresses. 


20000 


73946 


93946 


AS". 


two. 


=7/578 


22/09 


Tube. 
-3.0—1 


60792 


Wire  Envelope 


4.15 


dl  rest. 


17360 
31220 


Jacket. 
-4.85  — 


Fig.  9. 

STRESSES  IN  ACTION,  P0=  42,000  LBS.  PEE  SQ.  IN. 

(See  figures  10  and  11.) 

28.  The  stresses  in  action,  P0  =  42,000  Ibs.  per  sq.  in.,  in  the 
tube  and  jacket  are  the  same  as  were  obtained  when  the  wire 
was  wrapped  under  constant  tension  and  are  given  by  equations 
(55)  to  (58),  inclusive,  and  (63)  to  (66),  inclusive.  The  stresses 
in  the  wire  envelope  are  obtained  by  adding  to  the  stresses, 
system  at  rest,  those  due  to  the  interior  pressure  P0  =  42,000  Ibs. 
per  sq.  in.  acting  on  the  gun  considered  as  a  single  cylinder.  The 
latter  stresses  have  already  been  determined  and  used  in  equations 
(59)  to  (62),  inclusive,  relating  to  the  stresses  in  action  in  the 
wire  envelope,  wrapped  at  constant  tension,  due  to  P0  =  42,000 
Ibs.  per  sq.  in. 

Inner  Surface  of  the  Wire  Envelope.  —  The  tangential  stress  is 

+21,843  +  31,499  =  +53,342  Ibs.  per  sq.  in.  tension.    (87) 
The  radial  stress  is 
-22,109  -24,500  =  - 46,609  Ibs.  per  sq.  in.  compression.  (88) 


ELASTIC  STRENGTH   OF  WIRE-WRAPPED  GUNS 


35 


Tangential  Stresses. 
PQ  =  42000 


Fig.  10. 


Radial  Stresses 
=  42000 


Fig.  11. 


36  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

Outer  Surface  of  the  Wire  Envelope.  —  The  tangential  stress 
is 

+  45,344  +  16,615  =  +61,959  Ibs.  per  sq.  in.  tension.    (89) 

The  radial  stress  is 
-16,038  -  9616  =  -25,654  Ibs.  per  sq.  in.  compression.  (90) 

Figs.  10  and  11  show  graphically  the  tangential  and  radial 
stresses,  respectively,  system  in  action,  when  P0  =  42,000  Ibs. 
per  sq.  in. 

29.  Problem.  —  Find  for  a  6-inch  wire-wrapped  gun  whose 
parts  have  the  same  dimensions  and  elastic  limits  as  shown  in 
Fig.  7  and  whose  jacket  is  assembled  with  the  same  shrinkage, 
.007  inch,  but  on  the  tube  of  which  the  wire  is  wrapped  with  vary- 
ing tension  so  as  to  obtain  a  constant  tangential  stress  in  the  wire 
envelope,  system  in  action  under  the  required  powder  pressure: 

(a)  the  constant  tangential  stress  in  the  wire  envelope,  system  in 
action,  that  will  give  to  the  gun  the  same  tangential  elastic 
strength  as  when  the  wire  was  wrapped  under  a  constant 
tension  of  50,000  Ibs.  per  sq.  in.; 

(6)  the  tension  of  wrapping  at  the  inner  and  outer  surfaces  of  the 
wire  envelope; 

(c)  the  tangential  and  radial  stresses  at  the  inner  and  outer  sur- 

faces of  each  part,  system  in  action  under  the  maximum 
powder  pressure  which  the  gun  can  withstand  without  the 
tangential  stress  on  any  part  exceeding  the  elastic  limit  in 
tension  or  compression;  and 

(d)  the  tangential  and  radial  stresses  at  the  inner  and  outer  sur- 

faces of  each  part,  system  at  rest. 


CHAPTER  II. 

DETERMINATION  OF  THE  FORCES  BROUGHT  UPON  THE 
PRINCIPAL  PARTS  OF  THE  3-INCH  FIELD  CARRIAGE 
BY  THE  DISCHARGE  OF  THE  GUN. 

30.   Stability.  —  Total  Resistance  Opposed  to   Recoil  of  Gun. 

-  This  carriage  is  so  designed  that  when  the  gun  is  fired  at  0 
degrees  elevation  the  carriage  will  not  move  to  the  rear  and  the 
wheels  will  not  rise  from  the  ground.  Under  these  conditions 
the  carriage  is  said  to  be  stable.  To  prevent  the  carriage  from 
moving  to  the  rear  when  the  gun  is  fired,  there  is  provided  at 
the  end  of  the  trail  a  spade  of  such  an  area  that  the  horizontal 
resistance  which  the  ground  can  exert  against  it  is  greater  than 
the  sum  of  the  horizontal  components  of  the  forces  brought  upon 
the  carriage  by  the  discharge  of  the  gun;  and  to  prevent  the 
wheels  from  rising  from  the  ground  the  moment  of  the  weight  of 
the  system,  gun  and  carriage,  around  the  point  of  support  of  the 
trail  on  the  ground  is  made  greater  than  the  sum  of  the  moments 
about  that  point  of  the  forces  brought  upon  the  carriage  by  the 
discharge  of  the  gun.  The  forces  brought  upon  the  carriage  by 
the  discharge  of  the  gun  will  depend  on  the  resistance  opposed  to 
the  recoil  of  the  gun.  If  no  resistance  is  thus  opposed,  no  force 
will  be  brought  upon  the  carriage  by  the  discharge  of  the  gun 
except  the  slight  friction  between  the  gun  and  the  carriage  due 
to  the  weight  of  the  gun  and  its  movement  in  recoil.  The  weight 
of  the  gun  acts  on  the  carriage  at  all  times.  To  limit  the  recoil 
of  the  gun  a  resistance  must,  however,  be  imposed.  This  re- 
sistance is  imposed  through  the  hydraulic  brake  at  a  d  stance  of 
7.156  inches  below  the  axis  of  the  gun.  The  action  line  of  the 
resistance  being  below  the  axis  of  the  gun,  it  will  tend  to  rotate 
the  gun  around  its  center  of  mass,  assumed  to  be  in  that  axis. 
This  rotation  will  be  prevented  by  the  clips  on  the  gun  engaging 
with  those  on  the  cradle  (see  Fig.  15),  the  clip  on  the  cradle  exert- 
ing a  downward  force  on  the  forward  end  of  the  front  clip  of  the 

37 


38  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

gun  and  an  upward  force  on  the  clip  of  the  gun  at  the  rear  end 
of  the  cradle.  These  forces  will  produce  friction  as  the  gun 
recoils,  the  action  line  of  the  friction  being  parallel  to  the  axis  of 
the  gun  and  along  the  contact  surfaces  of  the  clips.  The  fric- 
tion also  opposes  the  recoil  of  the  gun.  The  direction  of  the 
motion  of  the  gun  due  to  the  action  of  the  powder  gases  is  in 
prolongation  of  its  axis,  and  since  this  direction  is  not  changed 
by  the  forces  exerted  on  the  gun  by  the  carriage,  it  follows  that 
the  resultant  of  all  the  latter  forces  must  be  a  force  whose  action  line 
coincides  with  the  axis  of  the  gun.  This  resultant  force  is  the  total 
resistance  which  opposes  the  recoil  of  the  gun. 

31.  Resultant  of  Forces  Exerted  by  the  Gun  on  the  Carriage. 
—  Since  the  forces  exerted  by  the  carriage  on  the  gun  are  like- 
wise exerted  by  the  gun  on  the  carriage,  but  in  opposite  directions, 
it  follows  that  the  resultant  of  all  the  forces  exerted  by  the  gun 
on  the  carriage  when  the  gun  is  fired  is  a  force  in  the  prolongation 
of  the  axis  of  the  gun  equal  and  opposite  in  direction  to  the  resistance 
which  opposes  the  recoil  of  the  gun.    As  the  carriage  remains  in 
equilibrium  when  the  gun  is  fired  the  forces  exerted  upon  it  by 
the  gun  develop  others  between  the  carriage  and  the  ground 
which  oppose  and  neutralize  the  effect  of  the  former  so  far  as 
motion  of  the  carriage  is  concerned.     These  forces  develop  others 
between  the  various  parts  of  the  carriage  which,  while  they  do 
not  cause  motion  of  the  parts  and,  therefore,  do  not  disturb  their 
equilibrium,  cause  stresses  in  the  materials  of  which  the  parts  are 
made,  which  stresses  must  not  exceed  safe  limits  in  order  that 
the  parts  of  the  carriage  may  not  be  distorted  or  broken.     The 
resultant  of  the  forces  exerted  by  the  gun  on  the  carriage,  which  is 
equal  and  opposite  in  direction  to  the  resistance  opposed  to  the 
recoil  of  the  gun,  is  represented  in  Fig.  12  by  R. 

32.  Limiting  Value  of  the  Resistance  R  Compatible  with  Sta- 
bility of  the  Carriage.  —  General  Expression  for  this  Resistance. 
(See  Fig.  12.)  —  Let  it  be  required  to  determine  the  greatest 
value  R'  which  the  force  R  may  have  without  causing  the  wheels 
of  the  carriage  to  rise  from  the  ground  when  the  gun  is  fired  at  0° 
elevation.     When  R  has  this  value  the  wheels  are  just  about  to 
rise  and  consequently  the  pressure  between  them  and  the  ground 
is  zero.     The  carriage  is  prevented  from  moving  to  the  rear 
under  the  action  of  the  force  R'  by  the  parallel  force  S  exerted  by 


DETERMINATION  OF  THE  FORCES 


39 


Vertical  through  C.G.  of 
\^     gun  in,  battery,  0 


40  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

the  ground  on  the  spade,  the  center  of  pressure  of  the  ground, 
and  therefore  the  point  of  application  of  the  force  S,  being  at 
a  distance  of  4  inches  below  the  surface. 

Since  the  pressure  between  the  wheels  and  the  ground  is  zero 
the  only  force  exerted  upward  on  the  carriage  by  the  ground  to 
counteract  the  force  of  gravity  on  the  system  will  be  the  vertical 
force  T  assumed  to  act  at  the  rear  point  of  contact  of  the  float  with 
the  ground,  the  point  about  which  moments  will  be  taken  to 
determine  the  conditions  of  stability.  The  forces  acting  on  the 
carriage  are,  therefore,  R'  acting  in  prolongation  of  the  axis  of 
the  gun;  the  weight,  2520  Ibs.,  of  the  system,  gun  and  carriage, 
acting  vertically  through  its  center  of  mass;  and  the  forces  S 
and  T  at  the  spade;  as  shown  in  Fig.  12.  As  the  carriage  is 
assumed  to  be  in  equilibrium  under  these  forces,  which  can, 
because  of  their  symmetrical  distribution,  be  considered  as  con- 
tained in  the  central  plane  through  the  axis  of  the  gun,  the  sum 
of  the  components  of  the  forces  in  the  vertical  and  horizontal 
directions  must  be  zero  and  the  sum  of  the  moments  of  the 
forces  about  any  point  in  their  plane  must  be  zero. 

Whence 

2520  -  T  =  0  or  T  =  2520  Ibs.  (1) 

S  -  R'  =  0    or    S  =  R'  (2) 

and  taking  moments  about  the  point  M  and  denoting  by  I  the 
lever  arm  of  the  weight  of  the  system  with  respect  to  this  point, 

R'  x  40.875  +5x4-  2520  X  I  =  0 

or  since  S  =  R'  from  equation  (2)' 

p,      2520  X  I 

44.875 

which  is  the  general  expression  for  the  greatest  value  of  the 
resistance  R  that  can  be  opposed  to  the  recoil  of  the  gun  when 
it  is  fired  at  0°  elevation  without  causing  the  wheels  to  rise  from 
the  ground. 

Limiting  Value  of  the  Resistance  Corresponding  to  I  =  105.8 
Ins.  —  Since  the  value  of  I  will  diminish  as  the  gun  recoils,  Rf  will 
have  a  maximum  value  when  the  gun  is  in  battery  and  a  minimum 
value  when  the  gun  is  at  its  extreme  limit  of  recoil;  and  it  will 


1NGL    - 


DETERMINATION  OF  THE  FORCES  41 

diminish  uniformly  with  the  distance  traveled  by  the  gun  in  recoil 
from  the  maximum  to  the  minimum  value.  The  value  of  I  when 
the  gun  is  in  battery  is,  from  Fig.  12,  105.8  ins.  and  the  corre- 
sponding value  of  R',  which  we  will  call  R\,  is 

2520  x  105.8      KQ/11  „ 

1  =  "     44875    •    =  M 

Limiting  Value  of  the  Resistance  Con-responding  to  I  =  88.66 
ins.  —  Limiting  Value  of  the  Resistance  Corresponding  to  a 
Length  of  Recoil  of  b  ft.  —  It  will  be  assumed  for  the  present 
that  the  length  of  recoil  of  the  gun  on  the  carriage  is  not  known. 
Since  the  value  of  R'  diminishes  uniformly  with  the  distance 
traveled  by  the  gun  in  recoil  its  value  R'z  corresponding  to  any 
distance  6  in  feet  recoiled  by  the  gun  can  be  determined  as  follows: 

Assume  any  distance  traveled  by  the  gun  in  recoil,  as  3.75  ft., 
and  compute  the  corresponding  position  of  the  center  of  gravity 
of  the  system  and  the  value  of  I.  This  has  already  been  done 
and  the  value  of  I  from  Fig.  12  is  88.66  ins.  The  value  of  R'z 
when  the  gun  has  recoiled  3.75  ft.  is,  therefore, 

D,       2520  x  88.66      AQrrQ  ,,  ,„ 

2  =  --  44875  -  =  ^ 

The  difference  between  R\  and  R'z  corresponding  to  a  distance 
of  3.75  ft.  recoiled  by  the  gun  is  962  Ibs.,  and,  therefore,  the 
difference  for  a  distance  of  b  ft.  recoiled  by  the  gun  is 

962 


3.75 


6  =  256.54  b  Ibs. 


The  value  of  R'2,  then,  corresponding  to  a  distance  of  b  ft.  re- 
coiled by  the  gun,  is 

R'i  =  R\  -  256.54  6  =  (5941  -  256.54  6)  Ibs.  (6) 

If  we  set  off  on  a  perpendicular  the  value  of  R\  to  any  scale 
from  equation  (4),  and,  on  a  second  perpendicular  at  a  horizontal 
distance  from  the  first  representing  to  any  scale  a  distance  of  3.75 
ft.,  the  value  of  R'z  from  equation  (5),  and  join  the  ends  of  the 
perpendiculars  by  a  right  line,  the  ordinates  of  this  line  will 
represent  the  limiting  values  of  R  corresponding  to  any  distance 
passed  over  by  the  gun  in  recoil  which  values  must  not  be  exceeded 


42  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

if  the  wheels  are  not  to  rise  from  the  ground  when  the  gun  is  fired 
at  an  elevation  of  zero  degrees.  This  line  is  shown  in  Fig.  13  by 
the  full  line. 

If  the  resistance  opposed  to  recoil  is  to  be  constant  it  may  have 
any  value  less  than  R'z,  but  in  order  to  meet  all  conditions  that 
may  arise  it  is  the  practice  to  make  it  at  all  points  of  recoil  sub- 
stantially less  than  R',  its  limiting  value. 


-.453- 


01  N. 

K  fc. 

Oi  O 

M-  *»• 

I  I 


Bj-0?      k 3'.75 -*       «« 

Fig.  13. 

Method  of  Varying  the  Resistance  Opposed  to  Recoil  Followed 
in  the  Design  of  this  Carriage.  —  In  the  design  of  the  carriage 
by  the  Ordnance  Department  the  resistance  is  made  constant 
during  the  time  the  powder  gases  are  acting  on  the  gun,  and  after 
they  cease  to  act  it  is  made  to  decrease  with  the  further  distance 
passed  over  in  recoil  as  shown  by  the  dot  and  dash  line  in  Fig.  13, 
the  rate  of  decrease  being  the  same  as  that  of  Rf,  its  limiting 
value.  This  arrangement  not  only  increases  the  margin  of  sta- 
bility of  the  carriage  during  the  early  stages  of  recoil,  as  com- 
pared with  a  constant  margin  of  stability,  but  facilitates  the 
calculation  of  the  value  of  the  resistance  corresponding  to  a  given 
length  of  recoil  of  the  gun  or  of  the  length  of  recoil  corresponding 
to  any  assumed  value  of  the  resistance. 

33.  Relation  Between  the  Varying  Values  of  the  Actual  Re- 
sistance and  the  Length  of  Recoil  of  the  Gun  on  the  Carriage.  — 
Determination  of  the  Values  of  RI,  R2,  and  b.  —  Let  V/  be  the 
maximum  velocity  of  free  recoil  of  the  gun  and  recoiling  parts 
determined  as  described  in  article  160,  page  275,  Lissak's  Ord- 
nance and  Gunnery,  r  the  time  corresponding  to  the  attainment  of 
Vf  in  free  recoil,  and  Ef  the  corresponding  space  that  would  have 
been  passed  over  by  the  gun  in  free  recoil  during  the  time  T;  T 
and  Ef  being  determined  as  described  in  article  163,  page  279, 
Lissak's  Ordnance  and  Gunnery. 


DETERMINATION  OF  THE  FORCES  43 

• 

Denoting  by  Ri  the  actual  resistance  to  recoil  during  the  time 
T  when  the  powder  gases  are  acting  on  the  gun,  and  making  it 
constant  during  this  time,  the  velocity  of  restrained  recoil  at 
the  end  of  the  time  T  will  be 


Mr  being  the  mass  of  the  gun  and  the  other  recoiling  parts,  cylin- 
der, oil,  counter-recoil  springs,  etc.;  and  R\/Mr  being  the  re- 
tardation of  these  parts  produced  by  the  resistance  R\. 

The  space  passed  over  by  the  gun  in  restrained  recoil  during 
the  time  T  will  be 


which  is  shown  in  Fig.  13  as  the  space  over  which  the  resistance 
to  the  recoil  of  the  gun  is  constant. 

Calling  Ri  the  value  of  the  resistance  when  the  gun  is  at  its 
extreme  limit  of  recoil  and  6  the  total  length  of  recoil  of  the  gun 
on  the  carriage  in  feet,  we  may  write 


which  expresses  the  fact  that  the  work  of  the  mean  resistance 
(Ri  +  Rz)/2  over  the  path  6  —  E^  is  equal  to  the  kinetic  energy 
of  the  recoiling  mass  at  the  end  of  the  time  T. 

Since  the  value  of  the  resistance  after  the  space  Err  has  been 
passed  over  by  the  gun  in  recoil  is  to  diminish  in  such  manner 
that  the  line  representing  its  diminishing  values,  Fig.  13,  is  to  be 
parallel  to  the  line  representing  the  diminishing  values  of  the 
limiting  resistance  Rf,  we  may  write  the  following  value  for  R2 
corresponding  to  a  total  distance  6  recoiled  by  the  gun  [see  de- 
duction of  equation  (6)]. 

R2  =  Rl  -  256.54  (6  -  #„)  (10) 

Substituting  this  value  of  R2  in  equation  (9)  and  the  values  of 
Err  and  Vrr  from  equations  (8)  and  (7),  respectively, 

\Ri  -  128.27  [b-Ef  +  (Ri/2  Mr)r2]H&  -Ef  +  (Ri/2  Mr)r2| 
=  [Mr/2]  [Vf  -  (Rt/MW  (11) 


44  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

In  equation  (11)  Ef,  Vf,  T,  and  Mr  are  known  so  that  only  RI 
and  6  are  unknown,  either  of  which  may  be  assumed  and  the  other 
determined.  Knowing  R:  and  b,  the  value  of  R2  may  be  obtained 
from  equation  (10)  by  substituting  therein  the  value  of  RI  and 
of  Err  from  equation  (8).  If  b  is  assumed  the  resulting  values  of 
RI  and  Rz  must  be  less  than  the  corresponding  values  of  the  limit- 
ing resistances  R\  and  R'2;  and  if  RI  is  assumed  b  must  not  be 
inconveniently  great  and  the  value  of  R2  must  be  less  than  that 
of  R'2.  In  the  ordinary  case  a  safe  value  for  RI  would  be  assumed 
and  the  corresponding  values  of  6  and  R2  calculated.  If  these 
values  were  satisfactory  they  would  be  accepted,  if  not  another 
value  for  RI  would  be  assumed  and  b  and  R2  recalculated;  or  if 
the  design  of  the  carriage  was  such  that  acceptable  values  of  RI, 
R2,  and  6  could  not  be  obtained  that  would  satisfy  equation  (11) 
the  design  would  have  to  be  changed. 

34.  Values  of  the  Actual  Resistance  to  Recoil  of  the  3-inch 
Field  Carriage.  —  Since  in  the  case  of  the  3-inch  field  carriage 
the  length  of  recoil  is  45  ins.,  equal  to  3.75  ft.,  a  value  of  6  =  3.75 
will  be  assumed  and  the  corresponding  values  of  RI  and  R2  cal- 
culated from  equations  (11)  and  (10).  For  this  carriage 


Vf  =  34.52  f.s.;  Ef  =  .48ft.;  r  =  .018  sees.;  Mr  =          =  29.813. 

oZ.Z 

Substituting  these  values  and  b  =  3.75  ft.  in  equation  (11)  and 
solving  for  RI  we  obtain 

Rt  =  4923  Ibs.  (12) 

Substituting  this  value  in  equation  (7) 


From  equation  (8) 


4923  x  (.018)2 


F        F  i     2  _   4o  .          _       q  f 

En  ~  Ef  ~  2M/  -  A*         2  x  29.813      "  A"    ft< 

From  equation  (10) 

z=Ri-  256.54  (6  -#„)  =  4923  -  256.54  (3.75  -  .453)  =  4077  Ibs. 

(15) 


DETERMINATION  OF  THE  FORCES  45 

The  actual  resistance  opposed  to  the  recoil  of  the  gun,  there- 
fore, has  a  value  of  4923  Ibs.  while  the  gun  is  recoiling  over  a  distance 
of  .453  ft.  and  then  decreases  uniformly  to  a  value  of  4077  Ibs. 
at  the  end  of  recoil,  the  length  of  recoil  being  3.75  ft.  or  45  ins. 
After  the  distance  of  .453  ft.  has  been  traveled  by  the  gun  in 
recoil,  the  equation  giving  the  resistance  at  any  point  is  as  follows: 

Rx  =  4923  -  256.54  (x  -  .453)  (16) 

in  which  x  is  the  distance  in  feet,  measured  from  the  origin  of 
movement,  over  which  the  gun  has  recoiled. 

Margin  of  Stability.  —  Referring  to  Fig.  13,  the  difference, 
called  the  margin  of  stability,  between  the  limiting  value  of  the 
resistance  that  may  be  opposed  to  the  recoil  of  the  gun  without 
causing  the  wheels  to  rise  from  the  ground,  and  the  actual  resist- 
ance employed  is  1018  Ibs.  at  the  beginning  of  recoil,  which  de- 
creases to  902  Ibs.  when  the  powder  gases  cease  to  act  on  the 
gun,  at  which  time  the  gun  has  recoiled  a  distance  of  .453  ft. 
Thereafter,  and  until  recoil  has  ceased  at  a  distance  of  3.75  ft. 
=  45  ins.,  the  margin  of  stability  remains  constant  and  equal  to 
902  Ibs. 

35.  Velocity  of  Restrained  Recoil  as  a  Function  of  Space.  — 
For  substitution  in  the  formula  which  gives  the  areas  of  the 
throttling  orifices  it  is  necessary  to  determine  the  velocity  of 
restrained  recoil  as  a  function  of  the  distance  recoiled.  Until 
the  gun  has  recoiled  a  distance  of  .453  ft.  the  resistance  is  constant 
and  the  velocity  of  restrained  recoil  as  a  function  of  space  must 
be  obtained  in  the  manner  described  in  article  166,  pages  283  and 
284,  Lissak's  Ordnance  and  Gunnery.  After  the  gun  has  recoiled 
a  distance  of  .453  ft.  the  velocity  of  restrained  recoil  correspond- 
ing to  any  distance  x  measured  from  the  origin  of  movement  is 
obtained  by  equating  the  work  remaining  to  be  done  by  the 
resistance  with  the  energy  of  the  recoiling  mass.  Thus,  calling 
Rx  the  value  of  the  resistance  at  the  point  x,  and  Vrx,  the  corre- 
sponding value  of  the  velocity  of  restrained  recoil, 

RX+RZ  fh  .    _  MrVrS 

~~  ~~~ 


or  R*  +  4077 


46  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

from  which  the  value  of  Vrx  corresponding  to  any  point  x  is 
obtained.  The  value  of  Rx  for  substitution  in  equation  (17)  is 
obtained  from  equation  (16). 

FORCES   ON  THE  CARRIAGE  CONSIDERED  AS  ONE  PIECE,   GUN 
AT  EXTREME  LIMIT  OF  RECOIL  AND  AT  15°  ELEVATION. 

36.  The  resistance  opposed  to  the  recoil  of  the  gun  on  the 
carriage  having  been  determined,  the  forces  on  the  carriage  con- 
sidered as  one  piece,  and  also  on  the  various  important  parts  of 
the  carriage  may  now  be  determined.  In  these  calculations  the 
gun  will  be  assumed  to  be  at  the  end  of  its  travel  in  recoil  after 
having  been  fired  at  15°  elevation.  For  a  complete  solution  of  the 
problem,  however,  the  position  of  the  gun  in  battery  should  be 
considered  and  other  conditions  of  elevation  also  assumed.  The 
parts  should  then  be  proportioned  to  resist  the  maximum  stresses 
brought  upon  them  under  any  of  the  assumed  conditions.  The 
method  of  calculating  the  forces  in  the  case  now  under  consider- 
ation is,  however,  exactly  like  that  to  be  followed  in  any  of  the 
other  cases  mentioned. 

The  gun  and  carriage  are  shown  diagrammatically  in  Fig.  14, 
the  gun  being  at  15°  elevation  and  at  its  extreme  limit  of  recoil. 
In  this  position  the  resistance  opposed  to  its  recoil  by  the  .carriage 
is  4077  Ibs.,  equation  (15),  which  is  equal  and  opposite  in  direction 
to  the  resultant  force  exerted  by  the  gun  on  the  carriage.  The 
latter  force  is  represented  in  position  and  direction  by  R2,  Fig.  14. 
The  horizontal  component  of  this  force  tends  to  move  the  carriage 
to  the  rear,  but  the  movement  is  prevented  by  the  horizontal 
force  S  exerted  by  the  ground  on  the  spade,  its  action  line  (center 
of  pressure)  being  at  a  distance  of  four  inches  below  the  surface 
of  the  ground.  The  vertical  component  of  the  force  R2  and  the 
weight  of  the  system  tend  to  produce  vertical  motion  down- 
ward, but  this  is  prevented  by  the  upward  pressure  of  the  ground 
at  L  and  at  T,  there  being  a  pressure  between  the  wheels  and  the 
ground  in  this  case  not  only  because  the  resistance  R2  is  less  than 
the  limiting  resistance,  but  also  because  the  gun  is  now  elevated 
above  zero  degrees.  The  forces  in  this  case  and  in  all  cases  to  be 
discussed  hereafter  can,  because  of  the  symmetrical  disposition 
of  the  parts,  be  considered  as  co-planar,  and  they  will  hereafter 
be  so  considered  without  further  reference  thereto. 


DETERMINATION  OF  THE  FORCES 


47 


48  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

Since  the  carriage  is  in  equilibrium,  the  sum  of  the  vertical 
components  of  the  forces  must  be  zero  as  must  the  sum  of  the 
horizontal  components,  and  also  the  sum  of  the  moments  of  the 
forces  about  any  point  in  their  plane.  Whence,  considering 
vertical  forces  acting  downward  ana  horizontal  forces  acting  in  a 
direction  opposite  to  that  of  recoil  as  positive, 

S  -4077 cos  15°  =  0  '(18) 

and 

4077  sin  15°  +  2520  -  L  -  T  =  0  (19) 

Taking  moments  about  /,  the  intersection  of  the  action  lines  of 
the  forces  R«  and  L,  and  considering  moments  that  tend  to  pro- 
duce clockwise  rotation  as  positive, 

2520  x  24.39  +  S  x  45.33  -  T  x  112.19  =  0          (20) 
From  equation  (18) 

S  =  4077  cos  15°  =  3938  Ibs.  (21) 

From  equation  (20) 

T  =  2139  Ibs.  (22) 

From  equation  (19) 

L  =  1436+  Ibs.  (23) 

FORCES  ON  THE  MOST  IMPORTANT  PARTS  OF  THE  CARRIAGE, 
GUN  AT  EXTREME  LIMIT  OF  RECOIL  AND  AT  15°  ELEVATION. 

37.  Forces  on  the  Gun.  —  Consider  first  the  gun  alone,  Fig. 
15.  The  resistance  opposed  to  its  recoil  is  4077  Ibs.  and  is  the 
resultant  of  a  number  of  forces  acting  between  the  carriage  and 
the  gun.  The  force  P  is  the  resistance  opposed  to  the  recoil  of 
the  gun  by  the  pressure  in  the  recoil  cylinder,  its  action  line  being 
parallel  to  the  axis  of  the  gun  and  at  a  distance  below  it  of  7.156 
ins.  As  it  is  not  exerted  through  the  center  of  mass  of  the  gun 
it  tends  to  rotate  it  about  that  center.  The  weight  of  the  gun, 
W0  —  835  Ibs.,  is  also  a  force  acting  on  it  as  shown,  but  as  it 
acts  through  the  center  of  mass  it  has  no  tendency  to  produce 
rotation.  The  rotation  of  the  gun  under  the  action  of  the  force 
P  is  prevented  by  the  engagement  of  the  clips  on  the  gun  with 
those  of  the  cradle  between  which  are  developed  the  forces  A 
acting  downward  on  the  gun  at  the  front  end  of  the  forward 


DETERMINATION  OF  THE  FORCES 


49 


gun  clip  and  B  acting  upward  at  the  rear  end  of  the  clip  on  the 
cradle.  The  forces  A  and  B  produce  friction  between  the  contact 
surfaces  of  the  clips  which  is  equal  to  /  x  (A  +  B},  f  being  the 
coefficient  of  friction  assumed  equal  to  .15.  This  friction,  which 


Fig.  15. 

will  be  called  F,  acts  along  the  contact  surfaces  of  the  clips  in  a 
direction  parallel  to  the  axis  of  the  gun  and  opposite  to  that  of 
recoil.  Its  action  line  may  be  taken  at  a  distance  of  3.65  ins. 
below  the  axis  of  the  gun  as  shown  in  Fig.  15,  and  since  it  also 
tends  to  produce  rotation  of  the  gun  around  its  center  of  mass, 
it  causes  additional  pressure  at  A  and  B  to  counteract  this  ten- 
dency. 

As  the  resultant  of  all  these  forces  is  a  resistance  of  4077  Ibs. 
acting  along  the  axis  of  the  gun  in  a  direction  opposite  to  that  of 
recoil,  we  may  write  three  equations  stating 

(a)  that  the  sum  of  the  components  of  all  the  forces  parallel  to 
the  axis  of  the  gun  is  equal  to  4077  Ibs.; 

(6)  that  the  sum  of  the  components  of  all  the  forces  in  a  direc- 
tion perpendicular  to  that  axis  is  zero,  since  the  resistance 
has  no  component  in  that  direction;  and 

(c)  that  the  sum  of  the  moments  of  the  forces  about  the  center  of 
mass  is  zero,  since  the  resistance  passes  through  that  center 
and,  therefore,  has  no  moment  with  respect  to  it. 

The  action  lines  and  location  of  all  the  forces  are  shown  hi 
Fig.  15.  Forces  acting  to  oppose  the  recoil  of  the  gun  and  those 
acting  downward  at  right  angles  to  the  axis  of  the  gun  are  con- 


50  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

sidered  positive.  Those  acting  in  opposite  directions  are  con- 
sidered negative.  Moments  that  tend  to  produce  clockwise 
rotation  are  considered  positive  and  those  tending  to  produce 
counter-clockwise  rotation,  negative.  Therefore, 

P  -  Wa  sin  15°  +  .15  (A  +  B)  =  4077  (24) 

A  -B  +  W0cosl5°  =0  (25) 

Px7.156+.15(A+B)x3.65  +  5xl0.1-Ax44.725  =  0    (26) 

Substituting  in  equations  (24)  and  (26)  the  value  of  B  obtained 
by  solving  equation  (25)  for  B  and  replacing  Wg  cos  15°  by  its 
value  835  cos  15°  =  807  Ibs.  and  Wg  sin  15°  by  its  value  of  216  Ibs. 
and  reducing,  we  have 

P  +  .3  A  =4172  (27) 

P  X  7.156  -  33.53  A  =  -8593  (28) 

Multiplying  both  members  of  equation  (27)  by  7.156  and  sub- 
tracting from  equation  (28) 

-35.6768  A  =  -38,448  (29) 

or  A  =  1078  Ibs.  (30) 

From  equation  (25)    B  =  1885  Ibs.  (31) 

From  equation  (27)  P  =  4172  -  .3  X  1078  =  3849  Ibs.  (32) 

F  =  .15  (A  +  B)  =  .15  (1078  +  1885)  =  444  Ibs.  (33) 

38.  Forces  on  the  Cradle.  —  The  forces  acting  between  the 
cradle  and  the  gun  when  the  latter  is  fired  act  on  the  cradle  in  a 
direction  opposite  to  that  in  which  they  act  on  the  gun.  They 
are  shown  in  position,  direction,  and  magnitude  in  Fig.  16.  Wc  = 
409  Ibs.,  the  weight  of  the  cradle,  acts  on  it  vertically  through  its 
center  of  mass.  The  forces  P,  F,  and  Wc  sin  15°  tend  to  move 
the  cradle  in  a  direction  parallel  to  that  of  the  recoil  of  the  gun, 
but  are  prevented  from  doing  so  by  the  pintle  of  the  cradle  bear- 
ing against  the  pintle  socket  of  the  rocker,  giving  rise  to  the  force 
C  between  the  contact  surfaces  which  acts  in  the  opposite  direc- 
tion but  parallel  to  the  force  P.  Rotation  of  the  cradle  about  its 
center  of  mass  under  the  action  of  the  forces  A,  B,  C,  and  F  is 
prevented  by  the  engagement  of  the  clips  of  the  pintle  on  the  cradle 
with  those  of  the  pintle  socket  of  the  rocker  and  by  the  pressure 


DETERMINATION  OF  THE  FOMC 


52  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

between  the  surfaces  of  the  rear  cradle  clip  and  the  part  of  the 
rocker  above  the  connection  for  the  elevating  screw,  giving  rise 
to  a  force  D  acting  downward  on  the  cradle  at  the  clips  of  the 
pintle  and  a  force  E  acting  upward  against  the  rear  cradle  clip, 
both  forces  being  perpendicular  to  the  axis  of  the  cradle.  The 
location  of  these  forces  is  shown  in  Fig.  16. 

As  the  cradle  is  in  equilibrium  under  the  forces  acting  upon 
it,  the  sum  of  all  the  components  of  the  forces  in  a  direction 
parallel  to  the  axis  of  the  cradle  must  be  zero  as  well  as  the  sum 
of  all  the  components  of  the  forces  in  a  direction  perpendicular 
to  that  axis.  The  sum  of  the  moments  of  the  forces  about  any 
point  in  their  plane  must  also  be  zero. 

Therefore,        C  -  444  -  3849  -  409  sin  15°  =  0  (34) 

1885  +  409  cos  15°  +  D  -  E  -  1078  =  0  (35) 

and  taking  moments  about  the  center  of  mass  of  the  cradle, 

1078  x  16.32  +  1885  x  18.305  +  444  x  3.506  +  C  x  3.939 

-  D  x  14.32  -Ex  10.80  =  0        (36) 

From  equation  (34)        C  =  4399  Ibs.  (37) 

Multiplying  both  members  of  equation  (35)  by  10.80  and  sub- 
tracting from  equation  (36) 

25.12  D  =  58,001.  (38) 

Whence  D  =  2309  Ibs.  (39) 

From  equation  (35)      E  =  3511  Ibs.  (40) 

39.  Forces  on  the  Rocker.  —  The  forces  acting  between  the 
cradle  and  the  rocker  act  on  the  latter  in  a  direction  opposite 
to  that  in  which  they  act  on  the  cradle.  They  are  shown  in 
position,  direction,  and  magnitude  in  Fig.  17.  Wr  =  56  Ibs., 
the  weight  of  the  rocker,  acts  on  it  vertically  through  its  center 
of  mass.  Movement  of  the  rocker  under  the  action  of  these 
forces  is  prevented  by  its  engagement  on  the  axle  and  by  the 
support  afforded  by  the  elevating  screw,  giving  rise  to  a  force 
between  the  axle  and  the  rocker  and  one  between  the  elevating 
screw  and  the  rocker.  The  direction  of  the  force  exerted  by 
the  axle  on  the  rocker  is  not  known  but  it  must  be  normal  to  the 
surfaces  in  contact  and  must,  therefore,  intersect  the  axis  of  the 


53 


54  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

axle.  This  force  may,  however,  be  resolved  into  two  components, 
one  horizontal  and  the  other  vertical,  and  when  the  intensities 
of  the  components  are  determined  the  direction  of  the  resultant 
can  be  obtained  it  desired.  These  components  are  shown  hi 
Fig.  17,  designated  as  G  and  H,  respectively.  Since  the  elevating 
screw  is  pivoted  both  to  the  rocker  and  the  trail  and  is,  therefore, 
free  to  rotate  about  its  connections  thereto,  the  action  line  of  the 
force  /,  Fig.  17,  exerted  by  it  on  the  rocker  must  coincide  with 
the  axis  of  the  screw,  which  is  vertical  when  the  gun  is  at  an  angle 
of  elevation  of  15°. 

As  the  rocker  is  in  equilibrium  under  the  forces  acting  on  it, 
we  may  write  three  equations  of  equilibrium  as  follows: 

G  -  4399  cos  15°  -  2309  sin  15°  +  3511  sin  15°  =  0       (41) 
3511  cos  15°  -  I  +  56  +  H  +  4399  sin  15°  -  2309  cos  15°  =  0  (42) 

Taking  moments  around  the  point  of  intersection  of  the  forces 
G  and  H, 

2309x1.5+4399x1.78+3511x23.62-7x21.55 

+56x5.65  =  0        (43) 

From  equation  (41)  G  =  3938  Ibs.  (44) 

From  equation  (43)  /  =  4387  Ibs.  (45) 

From  equation  (42)  H  =  2031  Ibs.  (46) 

40.  Forces  on  the  Axle,  Wheels,  Flasks,  and  Other  Parts 
Below  the  Rocker,  Considered  as  One  Piece.  —  The  forces 
acting  between  the  rocker  and  the  parts  below  it,  act  on  the  parts 
below  in  a  direction  opposite  to  that  in  which  they  act  on  the 
rocker.  They  are  shown  in  position,  direction,  and  magnitude 
hi  Fig.  18.  Wf  =  1220  Ibs.,  the  weight  of  all  the  parts  below  the 
rocker,  acts  on  them  through  their  center  of  mass.  All  move- 
ment of  the  parts  of  the  carriage  below  the  rocker  under  the  action 
of  these  forces  is  prevented  by  the  horizontal  and  vertical  reactions 
of  the  ground  which  give  rise  to  a  vertical  force  L  acting  upward 
on  the  wheel,  another  T  acting  upward  on  the  float  of  the  spade, 
and  a  horizontal  force  S  acting  against  the  spade  in  a  direction 
opposed  to  recoil.  T  is  assumed  to  act  on  the  float  at  its  rear 
point  of  contact  with  the  ground.  The  point  of  application  of  S 
is  at  the  center  of  pressure  of  the  earth  on  the  spade,  a  distance  of 


DETERMINATION  OF  THE  FORCES 


55 


56  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

four  inches  below  the  surface  of  the  ground.     The  forces  L,  T, 
and  S  are  shown  in  position,  direction,  and  magnitude  in  Fig.  18. 
Writing  the  ordinary  equations  of  equilibrium 

4387  -  T  +  1220  -  L  -  2031  =  0  (47) 

S  -  3938  =  0  (48) 

Taking  moments  about  the  center  of  the  axle,  the  point  of 
intersection  of  the  forces  G,  H,  and  L, 

4387  x  21.55  +  S  x  32.0  -  T  x  112.19  +  1220  X  15.938  =  0  (49) 
From  equation  (48)  S  =  3938  Ibs.  (50) 

From  equation  (49)  T  =  2139  Ibs.  (51) 

From  equation  (47)  L  =  1437  Ibs.  (52) 

Comparing  the  values  of  S,  T,  and  L  given  in  equations  (50), 
(51),  and  (52)  with  those  given  in  equations  (21),  (22),  and  (23) 
deduced  by  considering  the  carriage  as  one  piece,  it  will  be  seen 
that  the  values  of  S  and  T  deduced  by  the  two  methods  are 
identical  and  the  values  of  L  differ  by  less  than  one  pound,  which 
proves  the  correctness  of  the  values  of  all  the  forces  determined 
above.  The  very  slight  difference  in  the  values  of  L  deduced  by 
the  two  methods  is  due  to  not  carrying  out  the  values  of  the 
forces  on  the  various  parts  beyond  the  decimal  point. 

PROBLEMS. 

41.  Figs.  19  to  24,  inclusive,  indicate  diagrammatically  the 
location,  direction,  and  magnitude  of  the  forces  on  the  5-inch 
barbette  carriage,  model  1903,  when  the  gun  has  been  fired  at  0° 
elevation  and  is  at  its  extreme  limit  of  recoil.  This  carriage  is  in 
all  respects  similar  to  the  6-inch  barbette  carriage  described  in 
article  195,  pages  335  to  337,  Lissak's  Ordnance  and  Gunnery. 
It  was  designed  to  present  a  resistance  of  92,250  Ibs.  to  the  recoil 
of  the  gun. 


DETERMINATION  OF  THE  FORCES 


57 


1.  Assuming  a  resistance  of  92,250  Ibs.  to  the  recoil  of  the  gun, 
determine  the  values  of  the  forces  S,  T,  and  L  shown  in 
Fig.  19.  Explain  why  these  forces  are  developed  at  the 
points  indicated. 


58 


ST&ESSES  IN  GUNS  AND  GUN  CARRIAGES 


o 

c 

3 


Cn 


Ibs. 


^  R  = 


Ibs. 


?>   J 


CO 


f- 


+10.0+ 


+-I5.75-* 


A- 19485  Ibs. 


^ 
to 

Oi 

t^ 


C 
c 
p 


s 

lo 

§ 


2.  Assume  the  same  resistance  to  the  recoil  of  the  gun  as  in 
problem  1,  and  determine  the  values  of  the  forces  A,  B, 
P,  F,  and  F'  shown  in  Fig.  20.  Explain  why  these  forces  are 
developed  at  the  points  indicated. 


DETERMINATION  OF  THE  FORCES 

I 


59 


3.  Assume  the  values  of  the  forces  A,  B,  P,  F,  F',  and  Wc,  and 
determine  the  values  of  the  forces  C,  D,  and  E  shown  in 
Fig.  21.  Explain  why  these  forces  are  developed  at  the 
points  indicated. 


60 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


I 

bd 


I 

1 


I 


i 


i 

i 


4.  Assume  the  values  of  the  forces  C,  D,  E,  and 
WpV,  and  determine  the  values  of  the  forces 
G,  H,  and  /  shown  in  Fig.  22.  Explain  why 
these  forces  are  developed  at  the  points  in- 
dicated. 


ENGINE  Sil 
CANNON 


Nil  DOIIBAU 


DETERMINATION  OF  THE  FORCES 


61 


62 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


G-  155057   Ibs. 


24.65 *fc 


S- 92250   Ibs. 


Fig.  24.  —  Forces  on  the  Pedestal. 

5.  Assume  the  values  of  the  forces  G,  H,  I,  and  Wp,  and  deter- 
mine the  values  of  the  forces  S,  T,  and  L  shown  in  Fig.  24. 
Explain  why  these  forces  are  developed  at  the  points  in- 
dicated. 


CHAPTER  III. 

DETERMINATION  OF  THE  FORCES  BROUGHT  UPON  THE 
PRINCIPAL  PARTS  OF  A  DISAPPEARING  GUN  CARRIAGE 

BY   THE  DISCHARGE  OF  THE  GUN. 

^ 

42.  The  6-inch  disappearing  carriage  model  of  1905  Ml  is 
chosen  to  illustrate  the  subject;  and  the  forces  brought  upon 
the  gun,  gun  levers,  top  carriage,  and  elevating  arm  will  be  de- 
termined. 

The  first  step  is  to  obtain  the  curve  showing  the  velocity  of  free 
recoil  of  the  gun  as  a  function  of  time. 

Velocity  of  the  Projectile  in  the  Bore.  —  Fig.  25  shows  the 
velocity  of  the  projectile  in  the  bore  as  a  function  of  the  travel  of 
the  projectile  in  inches. 


Axis  of  travel. 
Fig.  25. 


The  ordinates  of  this  curve  corresponding  to  various  values  of 
the  travel  of  the  projectile  calculated  by  the  formulas  of  interior 
ballistics  are  given  in  Table  1. 

63 


64 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 
TABLE  1. 


Travel  of 
projectile  in 
inches. 

Velocity  in 
bore,  feet 
per  second. 

Travel  of 
projectile  in 
inches. 

Velocity  in 
bore,  feet 
per  second. 

2 

223.24 

54 

1776.61 

4 

411.60 

60 

1829.30 

6 

577.41 

72 

1924.05, 

8 

714.57 

84 

2004.32 

10 

835.79 

96 

2076.75 

12 

940.95 

108 

2142.55 

15 

1074.25 

120 

2202.73 

18 

1184.05 

132 

2259.86 

21 

1275.70 

144 

2309.20 

24 

1353.46 

156 

2356.67 

27 

1419.15 

168 

2400.89 

30 

1477.17 

180 

2442.23 

33 

1527.95 

192 

2481.01 

36 

1573.20 

204 

2517.49 

42 

1651.56 

216 

2551.89 

48 

1718.15 

234 

2600.02 

Curve  of  Reciprocals.  —  Velocity  and  Travel  of  Gun  in  Free 
Recoil  while  the  Projectile  is  in  the  Bore.  —  Fig.  26  shows  the 
reciprocals  of  the  velocities  of  the  projectile  in  the  bore  as  a  func- 
tion of  the  travel  of  the  projectile  in  inches. 


Axis  of  travel. 
Fig.  26. 

Table  2  gives  the  values  of  the  reciprocals  of  the  velocities  of 
the  projectile  in  the  bore  corresponding  to  various  values  of  the 
travel  of  the  projectile,  the  corresponding  times  obtained  by 


DETERMINATION  OF  THE  FORCES 


65 


measuring  the  areas  under  the  curve  of  Fig.  26  up  to  the  corre- 
sponding limiting  ordinates,  the  corresponding  values  of  the 
velocities  of  free  recoil  of  the  gun,  and  the  corresponding  distances 
traveled  by  it  in  free  recoil.  The  velocity  of  free  recoil  of  the  gun 
while  the  projectile  is  in  the  bore  is  obtained  from  equation  (2), 
page  275,  Lissak's  Ordnance  and  Gunnery,  which  is  as  follows: 


/!«  — 


w 


in  which  w,  the  weight  of  the  projectile,  is  106  Ibs.  ;  o>,  the  weight 
of  the  powder  charge,  is  30  Ibs.;  W,  the  weight  of  the  gun,  is 
12,764  Ibs.,  and  vf  is  the  velocity  of  free  recoil  of  the  gun  corre- 


TABLE  2. 


Travel  of 
projectile  in 
inches. 

Reciprocal  of 
velocity  in  bore. 

Time  in  seconds. 

Velocity  of  free 
recoil  of  gun, 
feet  per  second. 

Length  of  free 
recoil  in  feet. 

2 

.0044796 

.0014932 

2.116 

.00158 

4 

.0024296 

.0020187 

3.902 

.00316 

6 

.0017359 

.0023557 

5.474 

.00474 

8 

.0013995 

.0026157 

6.774 

.00632 

10 

.0011965 

.0028308 

7.923 

.00790 

12 

.0010628 

.0030184 

8.920 

.00948 

15 

.0009309 

.0032665 

10.181 

.01185 

18 

.0008446 

.0034879 

11.225 

.01422 

21 

.0007839 

.0036912 

12.094 

.01659 

24 

.0007389 

.0038814 

12.831 

.01896 

27 

.0007047 

.0040617 

13.454 

.02133 

30 

.0006770 

.0042343 

14.003 

.02370 

33 

.0006545 

.0044007 

14.485 

.02607 

36 

.0006356 

.0045619 

14.914 

.02844 

42 

.0006055 

.0048720 

15.657 

.03318 

48 

.0005820 

.0051687 

16.288 

.03792 

54 

.0005629 

.0054548 

16.842 

.04266 

60 

.0005467 

.0057196 

17.342 

.04740 

72 

.0005194 

.0062595 

18.240 

.05688 

84 

.0004989 

.0067616 

19.001 

.06636 

96 

.0004815 

.0072517 

19.688 

.07584 

108 

.0004667 

.0077257 

20.312 

.08532 

120 

.0004540 

.0081860 

20.882 

.09480 

132 

.0004425 

.0086342 

21.424 

.10428 

144 

.0004331 

.0090712 

21.892 

.11376 

156 

.0004243 

.0094994 

22.341 

.12324 

168 

.0004651 

.0099198 

22.760 

.13272 

180 

.0004095 

.010333 

23.152 

.14220 

192 

.0004031 

.010739 

23.520 

.15168 

204 

.0003972 

.011139 

23.866 

.16116 

216 

.0003919 

.011534 

24.192 

.17064 

234 

.0003846 

.012116 

24.648 

.18486 

T  =  .0606  second. 


E  =  1.6786  feet. 


66 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


spending  to  a  velocity  v  of  the  projectile.  The  distance  traveled 
by  the  gun  in  free  recoil  while  the  projectile  is  in  the  bore  is  given 
by  the  formula 

_  w  +  %<Zu 
~~W~T& 

hi  which  w,  o>,  and  W  have  the  same  meanings  as  before,  u  is  the 
travel  of  the  projectile  in  inches,  and  uf  is  the  distance  in  feet 
traveled  by  the  gun  in  free  recoil. 

Curve  of  Free  Recoil.  —  From  the  data  given  in  the  third  and 
fourth  cohimns  of  table  2  the  curve  showing  the  velocity  of  free 
recoil  of  the  gun  as  a  function  of  time  may  be  plotted  up  to  the 
time  when  the  projectile  leaves  the  bore  of  the  gun.  The  re- 
mainder of  this  curve  up  to  the  time  when  the  maximum  velocity 
of  free  recoil  is  reached  and  the  powder  gases  cease  to  act  on  the 
gun  is  obtained  as  outlined  in  paragraph  163,  page  278,  Lissak's 
Ordnance  and  Gunnery.  The  maximum  velocity  of  free  recoil 
computed  from  equation  (4),  page  275,  Lissak's  Ordnance  and 

Gunnery,  is 

Vf  =  (wV  +  4700w)/PF  =  32.64  f.s. 


Vf  =32.64 


t-  012116  seconds       Axis  of  time. 
Fig.  27. 


t  =.0606  seconds. 


With  this  value  as  an  ordinate  to  the  proper  scale  a  line  is 
drawn  parallel  to  the  axis  of  t  and  the  curve  obtained  from  table  2 
is  continued  with  its  general  rate  of  change  in  curvature  until  it 
becomes  tangent  to  this  line.  Fig.  27  shows  the  curve  so  ob- 
tained, representing  the  velocity  of  free  recoil  of  the  gun  as  a 
function  of  time. 


DETERMINATION  OF  THE  FORCES  67 

From  Fig.  27  the  time  when  the  maximum  velocity  of  free 
recoil  is  reached  is  .0606  sees.  This  time  is  designated  by  r. 

To  obtain  to  a  close  degree  of  approximation  the  distance 
traveled  by  the  gun  in  free  recoil  after  the  projectile  has  left  the 
bore  and  until  the  powder  gases  have  ceased  to  act,  the  velocities 
of  free  recoil  corresponding  to  .016,  .020,  .028,  .038,  and  .048  sees. 
are  measured  from  the  curve  shown  in  Fig.  27.  We  also  have 
.012116  and  .0606  as  the  times  when  the  projectile  left  the  bore 
and  when  the  powder  gases  ceased  to  act,  respectively.  The 
velocities  corresponding  to  these  times  in  sequence  are  24.648, 
27.103,  28.764,  30.715,  31.881,  32.442,  and  32.64  f.  s.,  respectively. 
Assuming  that  between  any  two  of  these  times  the  velocity  in- 
creases uniformly,  the  space  passed  over  by  the  gun  in  free  recoil 
is  equal  to  the  product  of  the  difference  in  time  by  the  mean  veloc- 
ity during  the  interval. 

For  the  first  interval 

24.648+27.103  _  ^        _ 


For  the  second  interval 
27.103+28.764 


and  similarly  the  spaces  passed  over  by  the  gun  during  each  of 
the  succeeding  intervals  are  found  to  be  .237196,  .31298,  .321615, 
and  .41001  ft.,  respectively.  The  sum  of  these  spaces,  1.49374  ft., 
is  the  distance  traveled  by  the  gun  in  free  recoil  after  the  projectile 
has  left  the  muzzle  and  until  the  powder  gases  have  ceased  to  act. 
Adding  to  this  distance  .18486  ft.,  the  space  over  which  the  gun 
traveled  in  free  recoil  while  the  projectile  was  in  the  bore,  we  have 
1.6786  ft.  as  the  total  distance  traveled  by  the  gun  in  free  recoil 
up  to  the  time  when  the  powder  gases  ceased  to  act  on  it.  This 
distance  is  designated  by  E. 

When  a  planimeter  is  available  the  distances  recoiled  may  be 
more  accurately  obtained  by  measuring  the  areas  under  the 
curve,  Fig.  27.  The  use  of  data  given  by  the  curve  of  free  recoil 
will  appear  later. 


68 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


43.  Length  and  Lettering  of  Parts.  —  Reference  to  Coordinate 
Axes.  —  The  parts  upon  which  the  forces  are  to  be  determined 
are  shown  in  Fig.  28. 


Fig.  28. 


DETERMINATION  OF  THE  FORCES 


69 


Fig.  29  is  a  diagram  showing  the  lengths  of  the  parts  under 
consideration  and  their  positions  with  respect  to  each  other  and 
to  the  axes  of  coordinates. 


Fig.  29. 

The  origin  of  coordinates  is  assumed  at  A,  the  projection  of 
the  axis  of  the  gun-lever  pin  when  the  gun  is  in  battery.  The 
forces  are  considered  as  co-planar  and  the  axis  of  X  will  be  taken 
as  horizontal  and  the  axis  of  Y  as  vertical.  ABC  is  the  median 
line  of  the  gun  lever,  a  is  the  length  of  the  part  between  the  axis 
of  the  gun-lever  pin  and  the  axis  of  the  gun-lever  axle,  and  0  is 


70 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


the  angle  which  this  part  makes  with  the  vertical;  c  is  the  length 
of  the  part  between  the  axes  of  the  gun-lever  axle  and  the  gun 
trunnions,  and  <f>  —  /3  is  the  angle  which  this  part  makes  with 
the  vertical.  /3  is  a  constant  angle.  CD  represents  the  axis  of 
the  gun,  the  points  C  and  D  representing  the  axes  of  the  gun 
trunnions  and  elevating-band  trunnions,  respectively;  d  is  the 
distance  between  these  points;  and  ^  is  the  angle  between  the 
axis  of  the  gun  and  the  horizontal.  At  the  instant  of  firing  ^ 
is  equal  to  the  angle  of  elevation.  ED  represents  the  elevating 
arm  pivoted  to  a  fixed  part  (the  elevating  slide  of  the  rear  tran- 
som) of  the  carriage  at  E  and  to  the  trunnions  of  the  elevating 
band  at  D;  and  6  is  the  angle  which  the  elevating  arm  makes 
with  the  vertical.  6  is  the  length  of  the  elevating  arm  and  I  the 
length  of  the  part  from  the  pivot  E  to  the  center  of  gravity  F  of 
the  arm.  x'T  and  y'r  are  the  coordinates  of  E.  The  gun-lever 
axle  is  carried  in  bearings  at  B  in  the  top  carriage,  which  during 
recoil  slides  along  the  roller  path  GI  inclined  at  an  angle  a  with 
the  horizontal.  The  path  of  the  axis  of  the  gun-lever  axle  during 
recoil  is  indicated  by  the  dot  and  dash  line  BK.  H  is  the  center 
of  mass  of  the  counterweight  directly  under  the  origin  of  co- 
ordinates, and  h  is  its  distance  below  that  point. 


Fig.  30. 

44.  Forces  on  the  Gun.  —  Fig.  30  represents  the  gun  and  the 
forces  acting  on  it  at  the  instant  when  the  maximum  powder 
pressure  occurs.  The  velocity  corresponding  to  the  point  of 
inflection  in  Fig.  27  is  10  f .  s.  which  is  therefore  the  velocity  of 
free  recoil  at  the  instant  of  maximum  powder  pressure.  From 
Table  2  the  corresponding  distance  traveled  by  the  gun  in  free 


DETERMINATION  OF  THE  FORCES  71 

recoil  is  .01151  ft.  =  .1381  ins.  The  actual  distance  recoiled  by 
the  gun  will  be  somewhat  less  than  this,  so  that  for  all  practical 
purposes  it  will  be  sufficiently  accurate  to  consider  the  coordinates 
of  the  various  points  of  the  system  and  the  dimensions  of  the  angles 
at  the  instant  when  the  maximum  powder  pressure  occurs  as  having 
the  same  values  as  they  had  when  the  gun  was  about  to  be  fired. 

At  the  time  of  maximum  powder  pressure  the  gases  will  exert 
a  force  F  on  the  gun  in  the  direction  of  its  axis.  The  weight  of 
the  gun  W0  is  a  force  acting  upon  it  through  its  center  of  mass 
which  is  assumed  to  be  in  the  axis  of  the  gun  trunnions.  The 
gun  levers  and  the  elevating  arm  restrain  the  free  movement  of 
the  gun,  retarding  it  in  the  direction  of  the  axis  of  X  and  giving 
it  an  acceleration  in  the  direction  of  the  axis  of  Y.  The  forces 
exerted  on  the  gun  by  these  parts  must  be  normal  to  the  surfaces 
in  contact  and  must,  therefore,  intersect  the  axes  of  the  gun 
trunnions  and  elevating-band  trunnions,  respectively,  but  as 
their  direction  is  not  known  the  force  which  the  gun  levers  exert 
on  the  gun  is  resolved  into  horizontal  and  vertical  components  P 
and  PI,  respectively,  and  the  force  exerted  on  the  gun  by  the  ele- 
vating arm  is  resolved  into  a  horizontal  component  P2  and  a 
vertical  component  P3. 

Since  the  forces  may  be  considered  as  co-planar  the  algebraic 
sum  of  their  horizontal  components  must  equal  the  product  of 
the  mass  of  the  gun  by  its  horizontal  acceleration  and  similarly 
for  the  vertical  components.  The  algebraic  sum  of  the  moments 
of  the  forces  with  respect  to  the  center  of  mass  must  also  equal  the 
moment  of  inertia  of  the  gun,  taken  with  respect  to  the  axis  through 
its  center  of  mass  perpendicular  to  the  plane  of  the  forces,  multi- 
plied by  the  angular  acceleration  about  that  axis.  We  may  then 
write  the  following  equations: 

F  cos  ^  -  P  -  Pz  =  Mgd*xg/dP  (1) 

-F  sin  t  -  Wg  -  P!  +  P3  =  M^/dt*  (2) 

sin  ^  -  P3d  cos  ^  =  2wrff2  x  d?t/dtz  (3) 


In  the  above  equations  the  subscript  g  indicates  that  the  sym- 
bols under  which  it  is  placed  refer  to  the  gun.  The  subscript  a 
will  be  similarly  used  for  the  gun  levers  and  the  subscripts  c,  e, 
and  w  for  the  top  carriage,  elevating  arm,  and  counterweight, 
respectively. 


72 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


45.  Forces  on  the  Gun  Levers.  —  Fig.  31  represents  the  gun 
levers  and  the  forces  acting  thereon. 

The  top  circle  represents  the  bearings  in  the  gun  levers  for  the 
gun  trunnions  to  which  the  upper  ends  of  the  levers  are  attached, 

y 


dt2 


Fig.  31. 


the  middle  circle  the  trunnions  of  the  gun-lever  axle  carried  in 
bearings  in  the  top  carriage,  and  the  bottom  circle  the  bearings 
for  the  gun-lever  pins  to  which  the  lower  ends  of  the  levers  are 
attached  and  to  which  are  also  fastened  the  counterweight 'and 


DETERMINATION  OF  THE  FORCES  73 

the  hydraulic  recoil  cylinder.  The  gun-lever  pins,  the  lower  ends 
of  the  gun  levers,  the  counterweight,  and  the  recoil  cylinder  are 
constrained  by  the  construction  of  the  carriage  to  move  only  in  a 
vertical  direction. 

The  forces  P  and  PI  which  the  gun  levers  exert  on  the  gun  are 
exerted  in  the  opposite  directions  as  shown  in  Fig.  31  by  the  gun 
on  the  gun  levers.  The  top  carriage  resists  the  translation  of  the 
gun  levers  and  the  force  exerted  by  it  upon  them  must  be  normal 
to  the  surfaces  in  contact  and  must,  therefore,  intersect  the  axis 
of  the  gun-lever  axle  trunnions.  As  the  direction  of  this  force  is 
not  known  it  will  be  resolved  into  horizontal  and  vertical  com- 
ponents P5  and  P4  as  shown  in  the  figure.  The  weight  of  the 
gun  levers  Wa  acts  at  the  center  of  mass  as  shown.  The  lower 
ends  of  the  levers  tend  to  move  to  the  front  around  the  gun-lever 
axle  under  the  action  of  the  forces  brought  upon  them  by  the  dis- 
charge of  the  gun,  but  are  prevented  from  doing  so  by  the  con- 
struction of  the  carriage  which  brings  a  force  P6  to  bear  upon 
them  as  shown.  The  force  P6  also  produces  a  force  of  friction  /P6 
between  the  vertical  guides  of  the  carriage  and  the  cross-head, 
which  is  attached  to  the  gun-lever  pins  and  forms  a  part  of  the 
counterweight.  This  friction  is  due  to  the  upward  movement  of 
the  lower  part  of  the  levers,  counterweight,  etc.  /  is  the  coeffi- 
cient of  sliding  friction.  The  force  R,  which  is  the  constant  resist- 
ance to  recoil  of  the  hydraulic  cylinder,  acts  vertically  on  the 
lower  ends  of  the  gun  levers  through  the  cross-head  and  gun-lever 
pins  as  does  the  weight  Ww  of  the  counterweight.  The  counter- 
weight must  also  be  given  an  acceleration  in  the  direction  of  the 
axis  of  Y  during  recoil  and  the  vertical  force  on  the  lower  ends  of 
the  gun  levers  necessary  for  this  purpose  is  measured  by  the 
product  of  the  mass  of  the  counterweight  by  its  acceleration. 
This  force  is  represented  by 


Placing  the  algebraic  sum  of  the  components  of  the  various 
forces  acting  on  the  gun  levers  in  the  direction  of  each  of  the 
two  axes  equal,  respectively,  to  the  product  of  the  mass  of  the 
gun  levers  and  its  acceleration  in  each  of  those  directions,  we  have 

•  P  -  Po  +  Pe  =  Mad*xa/dt*  (4) 

P!  +  P4  -  Wa  -  /P6  -  R  -  Wv  -  Mwd*yw/dP  =  Mad*ya/dP  (5) 


74 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


Placing  the  algebraic  sum  of  the  moments  of  the  forces  about 
the  center  of  mass  of  the  gun  levers  equal  to  the  moment  of  in- 
ertia of  the  gun  levers,  taken  with  respect  to  an  axis  through 
their  center  of  mass  perpendicular  to  the  plane  of  the  forces, 
multiplied  by  the  angular  acceleration  about  that  axis;  and 
representing  the  coordinates  of  the  centers  of  mass  of  the  gun, 
gun  levers,  and  gun-lever  pins  by  xg,  yg;  xa,  ya;  and  yp,  respec- 
tively, (xp  =  0)  we  have,  since  the  center  of  mass  of  the  gun 
levers  is  in  the  axis  of  their  trunnions, 


-  Va}    -  Pl(Xa   -  Xa}    - 

Ww  +  Mwd*yw/dP]xa  = 


(6) 


46.  Forces  on  the  Top  Carriage.  —  Fig.  32  represents  the  top 
carriage  and  the  forces  acting  thereon. 


,  L  8 


Fig.  32. 

The  horizontal  and  vertical  forces  P5  and  P4  which  the  top 
carriage  exerts  on  the  gun  levers  are  exerted  in  the  opposite 
directions  as  shown  in  Fig.  32  by  the  gun  levers  on  the  top  car- 
riage. The  top  carriage  rests  on  rollers  S,  and  for  the  purpose 
of  the  determination  of  the  forces  it  will  be  assumed  that  it  bears 
on  the  front  and  rear  rollers  only.  These  rollers  exert  forces  P^ 
and  P8  on  the  top  carriage  in  a  direction  normal  to  the  surface  of 
its  roller  path,  that  is,  upward  but  making  an  angle  a  with  the 
vertical.  These  forces  when  the  top  carriage  moves  to  the  rear 
give  rise  to  the  forces  of  friction  f'P7  and  f'P8  parallel  to  the 
surface  of  the  roller  path  and  passing  through  the  points  of  con- 
tact of  the  top  carriage  with  the  rollers.  /'  is  the  coefficient  of 
rolling  friction.  The  weight  Wc  of  the  top  carriage  acts  on  it 


DETERMINATION  OF  THE  FORCES 


75 


through  its  center  of  mass  F  as  shown.  The  lever  arms  of  the 
forces  P5,  P4,  P7,  P&,  /'P7,  and  /'P8  with  respect  to  the  center  of 
mass  are  li,  lz,  k,  k,  and  k,  respectively,  the  lever  arms  of  /'P7 
and  f'P8  being  equal. 

Placing  the  algebraic  sum  of  the  components  of  the  forces  in 
the  direction  of  each  of  the  two  axes  equal,  respectively,  to  the 
product  of  the  mass  of  the  top  carriage  and  its  acceleration  in 


Fig.  33. 

each  of  those  directions,  and  the  algebraic  sum  of  the  moments 
of  the  forces  about  the  center  of  mass  equal  to  zero,  since  there 
is  no  rotation  of  the  top  carriage,  we  have  the  following  equations: 
PB  -  P7  sin  a  -  Ps  sin  a  -  /' P7  cos  a  -  /'P8  cos  a  =Mcd2xc/dP  (7) 
P7  cos  a +P8  cos  a  -  P4  -f'P7  sin  a  -/'P8  sin  a  -  Wc  =  Mcd*yc/dt2  (8) 
P5Zi  -  P4k  +  P7Z3  -  PA  +  j'Pil*  +  /'  P*k  =  0  (9) 

47.  Forces  on  the  Elevating  Arm.  —  Fig.  33  represents  the 
elevating  arm  and  the  forces  acting  thereon. 


76  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

The  horizontal  and  vertical  forces  P2  and  P3,  which  the  elevat- 
ing arm  exerts  on  the  gun,  are  exerted  in  the  opposite  directions 
as  shown  in  the  figure  by  the  gun  on  the  elevating  arm.  A  pin 
through  the  lower  end  of  the  elevating  arm  rotates  in  bearings  in 
the  elevating  slide  which,  when  the  gun  is  being  elevated  or  de- 
pressed, slides  in  inclined  guideways  in  the  rear  transom  of  the 
carriage.  At  other  times  the  elevating  slide  is  stationary.  Mo- 
tion of  translation  of  the  pin  forming  the  lower  part  of  the  elevat- 
ing arm  under  the  action  of  the  weight  of  the  arm  and  the  forces 
P2  and  Ps  is  prevented  by  a  force  exerted  on  the  pin  by  the  elevat- 
ing slide.  This  force  is  normal  to  the  surfaces  in  contact  and, 
therefore,  intersects  the  axis  of  the  pin,  but  as  its  direction  is 
not  known  it  must  be  resolved  into  unknown  horizontal  and 
vertical  components  P9  and  P10  as  shown.  The  weight  of  the 
elevating  arm  acts  through  its  center  of  mass  F. 

Placing  the  algebraic  sum  of  the  components  of  these  forces  in 
the  direction  of  each  of  the  two  axes  equal,  respectively,  to  the 
product  of  the  mass  of  the  elevating  arm  and  its  acceleration  in 
each  of  those  directions;  and  the  algebraic  sum  of  the  moments 
of  the  forces  about  the  center  of  mass  equal  to  the  moment 
of  inertia  of  the  elevating  arm,  taken  with  respect  to  the  axis 
through  its  center  of  mass  perpendicular  to  the  plane  of  the 
forces,  multiplied  by  the  angular  acceleration  about  that  axis, 
we  have  p2  +  p9  =  Med2xe/dt2  (10) 

Pio  -  P3  -  We  =  Med?ye/dP  (11) 

P2  (6  -  1}  cos  0  +  P3  (b  -  0  sin  6  -  P9l  cos  6  +  P10l  sin  0 

=  2mreW8/dt-         (12) 

48.  Reduction  of  the  Number  of  Unknown  Quantities  in 
Equations  (1)  to  (12).  —  In  these  equations  F,  the  force  of  the 
powder  gases,  and  all  fixed  dimensions  are  known.  For  any 
assumed  position  of  the  gun  in  recoil  the  angles  0,  <£  —  ft,  \f/,  and 
6  as  well  as  the  coordinates  of  the  centers  of  mass  and  other 
important  points,  are  known  from  the  construction  of  the  carriage; 
and  further  it  is  sufficiently  accurate  for  all  practical  purposes 
to  assume  that  their  values  when  the  maximum  powder  pressure 
occurs  are  the  same  as  they  were  immediately  before  the  gun  was 
fired.  The  angles  a  and  ft  are  constant.  The  weights,  masses, 
and  moments  of  inertia  of  the  various  parts  are  also  known  from 


DETERMINATION  OF  THE  FORCES  77 

the  construction  of  the  carriage.  There  are,  therefore,  in  the 
twelve  equations  twenty-four  unknown  quantities  of  which 
twelve,  including  the  constant  resistance  of  the  recoil  cylinder,  are 
forces  and  twelve  are  accelerations,  linear  and  angular. 

There  are,  however,  certain  definite  relations,  true  at  all  times, 
between  the  centers  of  mass  and  other  important  points  of  the 
various  parts  of  the  carriage.  These  relations  may  be  expressed 
in  terms  of  the  trigonometrical  functions  of  the  variable  angles 
<£,  <t>  —  /?,  ty,  and  9;  and  from  them  by  differentiation  may  be  ob- 
tained similar  relations  between  the  various  velocities,  linear  and 
angular;  and  by  differentiating  again,  similar  relations  between 
the  accelerations,  linear  and  angular.  Having  obtained  these 
relations  between  the  accelerations,  all  of  them  may  be  expressed 
in  terms  of  dz(j>/dt-,  and  the  trigonometrical  functions  of  the 
variable  angles  which  are  known  for  any  assumed  position  of  the 
gun  in  recoil,  thus  reducing  the  unknown  quantities  in  equations 
(1)  to  (12),  inclusive,  to  one  angular  acceleration  and  twelve 
forces.  A  further  reduction  of  the  number  of  unknown  quantities 
to  twelve,  thereby  permitting  the  solution  of  the  equations  and 
the  determination  of  the  values  of  the  various  forces,  may  be 
made  by  obtaining  independently  the  value  of  R,  the  constant 
resistance  of  the  recoil  cylinder. 

49.  Method  of  Determining  R.  —  After  the  powder  gases 
have  ceased  to  act  on  the  gun,  the  work  of  the  resistance  R  over 
the  remainder  of  its  path  during  the  time  the  system  is  coming 
to  rest  plus  the  work  done  on  the  system  against  the  force  of 
gravity  during  this  time  is  equal  to  the  sum  of  the  kinetic  energies 
of  translation  and  rotation  possessed  by  the  system  at  the  begin- 
ning of  the  interval  of  time  considered.  Letting  x"  and  y"  with 
the  proper  subscripts  represent  the  coordinates  of  the  centers  of 
mass  of  the  various  parts  of  the  system  in  the  recoiled  position, 
and  x  and  y  with  the  proper  subscripts  the  coordinates  at  any  time 
during  recoil,  we  may  write  the  following  equation: 

R  (y"w-yw}+Ww  (y"w-yw}+Wa  (y"a-ya)+Wc(y"c-yJ 
-Wg  (yg-y"g)-We(ye-y"e)  =  (Mg/2)  (dxg/dt)* 
+  (Mg/2)  (dyg/dt?+  (Eror,V2)  (d+/dty+  (Ma/2)  (dxa/dt)*+ 
ya/dty  +  (Zmr<?/2)  (d4>/dt}z+(Mc/2}  (dxc/dt)z  * 
(dyc/dt}*+(Me/2}  (dxe/dt}*+(Me/2)  (d 
(dd/dt)2+(Mw/2)  (dyw/dt)z 


78  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

In  stating  equation  (13)  the  resistances  of  friction  have  been 
omitted  as  they  depend  largely  on  the  pressures  between  the 
parts  which  are  unknown.  As  a  result  the  value  of  R  deduced 
from  equation  (13),  while  sufficiently  exact,  will  be  somewhat 
larger  than  required  to  bring  the  system  to  rest  at  the  desired 
point. 

The  variable  coordinates,  yw,  ya,  etc.,  in  equation  (13)  may 
be  expressed  in  terms  of  the  trigonometrical  functions  of  the 
variable  angles,  and  the  differentials  may  be  expressed  in  terms 
of  these  functions  and  d<j>/dt.  The  variable  angles  may  also  be 
expressed  in  terms  of  <J>.  This  having  been  done  it  is  then  only 
necessary,  in  order  to  solve  equation  (13)  for  R,  to  select  some 
value  for  0  which  occurs  after  the  powder  gases  have  ceased  to 
act  on  the  gun,  and  to  determine  the  corresponding  value  of 
d<j>/dt.  The  data  given  by  the  curve  of  free  recoil  of  the  gun  will 
enable  the  required  values  of  <f>  and  d$/dt  to  be  chosen  with  suf- 
ficient accuracy,  and  the  value  of  </>  which  we  shall  select  is  that 
corresponding  to  the  instant  the  powder  gases  cease  to  act  on 
the  gun. 

50.  Value  of  6  and  of  the  Velocity  of  Restrained  Recoil  of  the 
Gun  at  the  Instant  the  Powder  Gases  Cease  to  Act  on  the  Gun.  — 
From  Table  2  it  is  seen  that  the  distance  traveled  by  the  gun  in 
free  recoil  up  to  the  time  when  the  projectile  leaves  the  bore  is 
.18486  ft.  and  the  corresponding  velocity  of  free  recoil  is  24.648 
f .  s.  The  value  of  xa  when  the  gun  is  in  battery  is  x'g  =  1.9628  ft. 
and,  assuming 

xg  -  x'g  =  .18486 

whence  xg  =  .1849  +  1.9628  =  2.1477, 

the  corresponding  value  of  </>  in  free  recoil  from  equation  (14) 
below  is  approximately  14°  9'.4.  Also  the  distance  traveled  by 
the  gun  in  free  recoil  up  to  the  time  ?  when  the  powder  gases 
cease  to  act  upon  it  is  1.6786  ft.,  the  corresponding  maximum 
velocity  of  free  recoil  is  32.64  f.  s.  and,  assuming 

xg  =  1.6786  +  x'g, 

the  corresponding  value  of  <£  in  free  recoil  is  approximately  23° 
37'.8. 

A  number  of  foreign  ordnance  engineers  of  prominence  assume, 
in  solving  problems  of  a  nature  similar  to  this,  that  the  maximum 


DETERMINATION  OF  THE  FORCES  79 

velocity  of  restrained  recoil  is  instantly  acquired  and  equal  to 
the  maximum  velocity  of  free  recoil;  but  the  experience  of  the 
Ordnance  Department,  U.  S.  Army,  has  shown  that  a  closer 
approximation  than  this,  and  one  sufficiently  exact  for  all  prac- 
tical purposes,  is  to  assume  that  due  to  the  restraint  of  the  recoil 
brake  the  velocity  of  the  gun  in  restrained  recoil  at  the  instant 
the  powder  gases  cease  to  act  on  it  is,  for  carriages  of  this  type, 
about  eight-tenths  of  the  maximum  velocity  of  free  recoil,  and 
that  the  value  of  the  angle  0  at  the  same  instant  is  equal  to  its 
value  when  the  gun  is  in  battery  plus  about  eight-tenths  of  its 
increase  in  value  in  free  recoil  up  to  the  time  r.*  The  value  of 
0  when  the  gun  is  in  battery  being  13°,  its  value  when  the  powder 
gases  cease  to  act  will  be  assumed  as  21°.  The  assumed  values 
of  <f>  and  VT  lie  between  those  obtained  above  for  free  recoil  at  the 
instant  the  projectile  leaves  the  bore  and  at  the  instant  the 
powder  gases  cease  to  act  on  the  gun,  respectively. 

51.  Determination  of  the  Values  of  the  Coordinates  of  the 
Centers  of  Mass  in  Terms  of  the  Trigonometrical  Functions  of 
the  Angles  0,  0  —  ft,  $,  6,  and  the  Constant  Angle  a.  —  Letting  x' 
and  y'  with  the  proper  subscripts  represent  the  coordinates  of 
the  centers  of  mass  of  the  various  parts  when  the  gun  is  in  bat- 
tery, and  x'r  and  y'r  the  coordinates  of  the  axis  of  the  pin  con- 
necting the  lower  end  of  the  elevating  arm  to  the  elevating  slide, 
we  may  write  the  following  equations  from  the  relations  shown 
in  Fig.  29: 

xa  =  a  sin  <j>  +  c  sin  (0  —  /3)  (14) 

xa  =  a  sin  0  (15) 

xc  =  a  sin  0  +  x'c  —  x'a  (16) 

xe  =  x'r  +  I  sin  0  (17) 
£e  =  xg  +  d  cos  ty  —  (b  —  1)  sin  6  =  a  sin  0  +  c  sin  (0  —  0) 

+  d  cos  $  -  (b  -  I)  sin  0  (18) 

Ma  —  y'a  +  (z<»  ~  z'a)  tan  a  +  c  cos  (0  —  /3)  =  y'a  +  a  sin  0  tan  a 

—x'a  tan  a  +  c  cos  (0  —  0)         (19) 

*  A  more  rigorous  method  of  determining  the  velocity  of  the  gun  in  re- 
strained recoil  and  the  value  of  <j>  at  the  instant  the  powder  gases  cease  to  act 
on  the  gun  is  used  by  the  gun-carriage  division  of  the  Ordnance  Department. 
It  is  too  extensive  for  discussion  here,  but  the  results  obtained  by  it  have  shown 
that  the  approximate  method  outlined  above  is  sufficiently  accurate  for  all 
practical  purposes. 


80  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

ya  -  y'a  +  (as.  -  z'u)  tan  a.  =  y'a  +  a  sin  0  tan  a  -  x'a  tan  a  (20) 

yc  =  y'a  +  a  sin  4,  tan  a  -  z'a  tan  a  -  (y'a  -  y'c}  (21) 

#P  =  y'a  +  a  sin  0  tan  a  —  x'0  tan  a  —  a  cos  0  (21?) 

2/w  =  2/'a  +  a  sin  ^  tan  a  —  x'a  tan  a  —  a  cos  0  —  h  (22) 

2/e  =  2/'r  +  Zcos0  (23) 

2/e  =  yg  —  d  sin  ^  —  (6  —  Z)  cos  6  =  y'a  +  a  sin  0  tan  «  —  x'a  tan  « 

+  c  cos  (0  -  j8)  -  d  sin  ^  -  (b  -  I)  cos  0  (24) 

52.  Determination  of  the  Values  of  ^  and  d  when  the  Value  of  0 
is  Known  or  Assumed.  —  Equating  the  values  of  xe  from  equations 
(17)  and  (18)  we  have 

6  sin  0  =  a  sin  0  +  c  sin  (<j>  —  /3)  +  d  cos  ^  —  x'r 
and  representing     a  sin  <j>  +  c  sin  (0  —  /3)  —  x'r  by  Z, 

6  sin  0  =  d  cos  $  +  Z  (25) 

Equating  the  values  of  ye  from  equations  (23)  and  (24)  we 
have 

6  cos  6  =  yfa  -f  a  sin  0  tan  a  —  x'a  tan  a  +  c  cos  (0  —  0) 

—  d  sin  ^  —  2/'r 
and  representing 

y'a  +  a  sin  0  tan  a  —  x'a  tan  a  +  c  cos  (0  —  /3)  —  2/'r  by  Y, 

6cos0  =  -dsini/'  +  7  (26) 

Squaring  equations  (25)  and  (26)  and  adding  we  have 

+  Z2  +  Y2 


which  may  be  put  in  the  form 

ZcosiA  -  YsinrA  =  (&2  -  d2  -  Zn~  - 


Dividing  by  VZ*  +  Y2 

Z  Y  b~  —  d2  —  Z-  —  Y2 


Let  A  =  tan-1  Z/Y 

whence  sin  A  =     ,_ 

VZ2  +  Y2 

and  cosA  =  V/+P 


BHWNEERINfl  :>JIiEAiJ 

CANNON   Sh 


DETERMINATION  OF  THE  FORCES  81 


Replacing  in  equation  (26^) 


by  sin  A  and  by  cos  A 


we  have 
sin  A  cos  »A  -  cos  A  sin  ^  =  sin  (A  -  £)  =     2  d  VZ2  +  Y2       (27) 


in  which  A  =  tan-1  Z/Y. 

For  any  value  of  0  the  corresponding  value  of  \f/  can  be  obtained 
from  equation  (27),  and  having  0  and  ^  the  corresponding  value 
of  6  can  be  obtained  from  equation  (26). 

53.  Determination  of  the  Velocities  in  Terms  of  the  Trigono- 
metrical Functions  of  the  Angles,  and  of  d$/dt.  —  Differentiating 
both  sides  of  equations  (14)  to  (24),  inclusive,  excepting  equation 
(21|),  and  dividing  by  dt  we  have 

dxg/dt  =  a  cos  0  (d<t>/dt)  +  c  cos  (0  -  0)  (d<}>/dt)  (28) 

dxa/dt  =  a  cos  0  (d<f>/dt)  (29) 

dxc/dt  =  a  cos  <£  (d$/dt)  (30) 

=  Z  cos  0  (d0/<ft)  (31) 
=  a  cos  <£  (dfy/dt)  +  c  cos  (</>  —  /3)  (d<l>/df)  — 

d  sin  iA  (dt/dt)  -  (b  -  1)  cos  0  (dfl/dO  (32) 

dya/dt  =  a  tan  a  cos  <£  (d<t>/dt)  -  c  sin  (0  -  0)  (d^/df)  (33) 

dya/dt  =  a  tan  a  cos  <£  (d$/d£)  (34) 

dyc/dt  =  a  tan  a  cos  0  (d<t>/dt)  (35) 

dyw/dt  =  a  tan  «  cos  0  (dfy/dt}  +  a  sin  0  (d^/dt)  (36) 

d^/e/^  =  -Zsin  0  (d8/dt)  (37) 
dye/dt  =  a  tan  a  cos  0  (d$/dt}  —  c  sin  (</>  —  /3)  (d<t>/d£) 

-  d  cos  ^  (dt/dt)  +  (b  -1}  sin  5  (de/dt}  (38) 

Equating  the  values  of  dxe/dt  given  by  equations  (31)  and 
(32)  we  have 

6  cos  0  (de/dt}  =  o  cos  0  (d$/dt}  +  c  cos  (0  -  0)  (d<j>/dt)  - 

d  sin  ^  (dt/dt)        (39) 

and  equating  the  values  of  dye/dt  given  by  equations  (37)  and 
(38)  we  have 

-6  sin  0  (de/df}  =  a  tan  a  cos  0  (d<t>/dt)  —c  sin  (0  —  0)  (d<t>/dt} 

-  d  cos  I  (d^/dt)        (40) 


82  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

Multiplying  equation  (39)  by  sin  0  and  equation  (40)  by  cos  0 
and  adding  we  have 

sin  e  a  cos  0  (d^/df)  +  sin  8  c  cos  (0  —  /3)  (d<j>/dt} 

+cos  e  a  tan  a  cos  0  (d0/dZ)  —  cos  0  c  sin  (0  —  /3)  (d<t>/dt) 
=  sin  0  d  sin  ^  (d^/dt)  +  cos  0  d  cos  ^  (d\l//dt) 


or 


sin  0  fa  cos  0  +  c  cos  (0  — 


d  (sin  0  sin  \b  +  cos  0  cos  \1/}  dd> 

—  ( 4  J_ ) 

cos0  fa  tan  a  cos  0  —  csin  (0  —  /3)|   J' 


d  (sin  0  sin  ^  -f-  cos  0  cos  i/O 

Dividing  equation  (39)  by  6  cos  0  we  have 

d0      a  cos  0  +  c  cos  (0  —  jS)  d0      d  sin 


d£  ~  6  cos  d  dt      b  cos  0  dt 

or  replacing  d^/dt  by  its  value  in  terms  of  d<j)/dt  from  equation 
(41)  and  representing  the  coefficient  of  d$/dt  in  that  equation 
by* 

dd  _  (a  cos  0  +  c  cos  (0  —  ff)  _  dsin^  ,J  d0  ,.„. 

~di~(  bcosd  ~  bcosO    }~dt 

For  any  value  of  d$/dt  the  corresponding  values  of  d^/dt 
and  dQ/dt  can  be  obtained  from  equations  (41)  and  (42),  respec- 
tively, when  the  angle  0  is  known.  The  values  of  all  the  linear 
velocities  can  then  be  obtained  from  equations  (28)  to  (37),  in- 
clusive. 

54.  Determination  of  the  Accelerations  in  Terms  of  the  Trigo- 
nometrical Functions  of  the  Angles,  and  of  d20/dt2.  —  Differ- 
entiating both  sides  of  equations  (28)  to  (38),  inclusive,  and 
dividing  by  dt  we  obtain 


dzx  /dtz  =  fa  cos  0  +  c  cos  (0  -  /3)  |  (<P$/dP)  —  fa  sin  0 

+  c  sin  (0  -  |8)  }  (d0/dZ)2  (43) 

dzxa/dP  =  a  cos  0  (d20/d<2)  -  a  sin  0  (d0/d*)2  (44) 

d?xc/dt2  =  a  cos  0  (d20/<&2)  -  a  sin  0  (d0/dZ)2  (45) 

=  lcosd  (d*e/dt2)  -  I  sin  6  (dB/dt?  (46) 


DETERMINATION  OF  THE  FORCES 


83 


d^/dt2  =  f  a  cos  0  +  c  cos  (0  -  0)  I  (d20/d*2) 

-  d  sin  ^  (dV/cfe2)  -  (b  -  I)  cos  0  (d2e/dt}2 

-  f  a  sin  0  +  c  sin  (0  -  /3) }  (d<j>/d£)2 

—  d  cos  ^  (d^/di)2  +  (b  —  I)  sin  0  (dd/dt}2 


(47) 


=  {a  tan  a  cos  0  -  c  sin  (0  -  0) }  (d2<t>/dt2) 

—  fa  tan  «  sin  0  +  c  cos  (0  -  0)  f  (d<f>/dt)2  (48) 

=  a  tan  «  cos  0  (d20/d*2)  -  a  tan  «  sin  0(d0/o'02  (49) 

d~yc/dt2  =  a  tan  a  cos  0  (d20/d£2)  -  a  tan  a  sin  0  (d0X^)2  (50) 

d^w/dl2  =  fa  tan  a  cos  0  +  a  sin  0  |  (d2$/dt2)  —  fa  tan  «  sin  0 

-  a  cos  0}  (d0/d02  (51) 

d-ye/dt2  =  - 1  sin  0  (d20/d*2)  -  Z  cos  0  (de/df)2  (52) 


(53) 


=  fa  tan  a  cos  0  -  c  sin  (0  -  0) }  ( 
-dcost  (d^/dtz)  +  (6  -  0  sin  0  (d2e/dt2) 
—  fa  tan  a  sin  0  +  c  cos  (0  —  /3) }  (d<t>/d£)2 
+  d  sin  ^  (dt/dt)2  +(b  -I)  cos  0  (dd/dty 


Equating  the  values  of  d2xe/dt2  from  equations  (46)  and  (47)  we 
have 

6  cos  0  (d2d/dt2}  -  b  sin  0  (dd/dt)2 

(54) 


=  f  a  cos  0  +  c  cos  (0  -  |8)}  (d20/d*2)  -  d  sin 

-  f  a  sin  0  +  c  sin  (0  —  0)  J  (d0/d02  -  d  cos  ^  (d^/dt)2 


and  equating  the  values  of  dzye/dt2  from  equations  (52)  and  (53) 

-6  sin  0  (d26/dt2)  -  b  cos  0  (dS/dt}2  = 

fa  tan  «  cos  0  -  c  sin  (0  -  /3)  |  (d2<t>/dt2}  -  d  cos  <KdV/<&2)     (55) 
—  {a  tan  a  sin  0  +  c  cos  (0  -  /3) }  (d0/d£)2  +  d  sin  ^  (d^/dty 

Multiplying  equation  (54)  by  sin  0  and  equation  (55)  by  cos  0 
and  adding  we  have 

-b(dB/dt}2  =  [sin  0  fa  cos  0  +  c  cos  (0  -  /3)  | 
+  cos  0  fa  tan  a  cos  0  -  c  sin  (0  -  18)  f]  (d20/a^2) 

-  [d  sin  0  sin  ^  +  d  cos  0  cos  ^]  (dV/d£2) 

—  [sin  0fasin0  +  csin  (0— /3)|  +cos  0  fa  tanasin0 
+  c  cos  (0  —  )8) }  ]  (d0/a^)2  -  [d  sin  0  cos  ^ 

-  d  cos  0  sin  ^]  (d^/dt)2 


84 
or 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


sin  6  \a  cos  <ft  +  c  cos  Oft  —  ff)  |  cfoft 
d  sin  6  sin  ^  +  d  cos  6  cos  ^     d£* 
cos  0  \a  tan  a  cos  <ft  —  c  sin  (<ft  —  j8) 


sm 


d  sin  0  sin  ^  +  d  cos  0  cos  ^ 
a  sin  <j>  -f-  c  sin  (<£  —  /3)  ( 


d2j> 
dt2 


d  sin  6  sin  ^  +  d  cos  6  cos 
cos  0  (a  tan  a  sin  <ft  +  c  cos  (<ft  —  ff) 

d  sin  6  sin  ^  +  d  cos  0  cos  ^ 
sin  d  cos  ^  —  cos  6  sin  ^  /d^ 
sin  6  sin  ^  +  cos  6  cos  ^ \d£ 


(56)' 


-       -- 

d  sin  0  sin  ^  +  d  cos  d  cos 

Dividing  equation  (54)  by  6  cos  0  and  solving  for  d20/dtz  we 
have 

a  cos  <ft  +  c  cos  (<fr  -  0)  dV 
6  cos  0  dt2 

_  d  sin  \f/  d2\f/ 
~  bcosd~di* 

_  a  sin  <f>  +  c  sin  (</>  —  /3)  /d</>\2 
6cos0  \dt) 


or  representing  the  coefficients  of 

d^/dP,     (d<t>/dt?,     (dt/dty,  and  (de/dt)2 

in  equation  (56)  by  X,  X',  X",  and  X'",  respectively,  we  have 


s-j 

6  cos 

a  sin  <ft  +  c  sin  (<ft  - 

6                         dt2 
-  0)  —  d  sin  \f/X'  /d<ft\2 

6  cos 
d  (cos  ^  —  sin  ^X'' 

0                        \d«/ 
)  /d^2 

d  sin  \{/X"f  —  b  sin 

\d«/ 

!/*V 

6  cos  6 

v^y                 j 

(57) 


*  When  this  equation  is  used  to  determine  the  value  of  dty/dP  the  numerical 
work  can  be  simplified  somewhat  by  noting  that  sin  8  sin  ^  +  cos  6  cos  ^  = 
cos  (0  —  ^)  and  sin  0  cos  ^  —  cos  e  sin  ^  =  sin  (0  —  \!/). 


DETERMINATION  OF  THE  FORCES  85 


For  any  value  of  d^/dP  the  corresponding  values  of 
and  d^d/dP  can  be  obtained  from  equations  (56)  and  (57),  re- 
spectively, when  the  angle  0  and  the  angular  velocities  are  known. 
The  values  of  the  linear  accelerations  can  then  be  obtained  from 
equations  (43)  to  (53),  inclusive. 

55.  Determination  of  the  Value  of  R  from  Equation  (13) 
(Gun  fired  at  0°  Elevation).  —  The  following  data  are  known 
from  the  construction  of  the  gun  and  carriage: 

TABLE  3. 

«  =  1°  20'  y'r  =  0.5563  ft.  Wg  =  12764  Ibs. 

0'  =  13°  y'g  =  9.2942  '  Mg  =  396.88 

0"  =  88°  y"g  =  4.6612  "  Wa  =  6040.8  Ibs. 

ft  =  2°  y'a  =  4.2224  "  Ma  =  187.84 

(0'  -  ft  =  11°  y"a  =  4.3008  "  We  =  2528  Ibs. 

(0"  -  |8)  =  86°  y'e  =  3.7571  "  Mc  =  78.61 

6'  =  3°  58'  y"c  =  3.8355  "  We  =  707  Ibs. 

f  =  angle  of  elevation  y'e  =  4.6548  "  Me  =  21.984 

y  =  5°  y"e  =  2.2571  "  Ww  =  19797  Ibs. 

x'g  =    1.9628ft.  y'p  =  0.0  Mw  =  615.58 

x"g  =    9.4857  "  y'v  =  -3.0    "  2mr2ff  =  14967 

x'a  =    0.975    "  y"w  =  1.1492  "  2mr2a  =  1250.4 

x"a  =    4.3317  "  a  =  4.3333  "  2wr2e  =  155.08 

x'e  =    1.6401  "  6  =  8.75      "  h  =  0.4653  ft. 

x"c  =    4.9968  "  c  =  5.1667  "  12  =  0.6651  " 

x'e  =    7.2145 "  d  =  5.5  13  =  2.4875  " 

x"e  =  10.579   "  I  =  4.1042  "  k  =  1.6792  " 

x'r  =    7.0002  "  fe  =  3.0  k  =  0.4255  " 

In  the  computation  of  R  it  is  customary  to  assume  that  the 
gun  is  fired  at  zero  degrees  elevation,  whence  ^  =  0,  and  this 
assumption  will  be  made  here.  No  difficulty  arises,  however, 
in  computing  R  for  any  other  given  elevation  of  the  gun  when 
fired,  the  method  to  be  pursued  in  that  case  being  exactly  similar 
to  that  which  follows. 

For  the  reasons  given  in  article  50,  page  78,  the  value  of  0  at 
the  instant  the  powder  gases  cease  to  act  on  the  gun  will  be 
taken  as  21°  and  the  corresponding  velocity  of  the  gun  in  re- 
strained recoil  as 

V~(dxg/dW  +  (dyg/dt)2  =  .8  Vf  =  26.112  f.s., 
Vf  being  the  maximum  velocity  of  free  recoil  equal  to  32.64  f.  s. 


86  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

With  <t>  =  21°  the  values  of  \f/  and  0,  computed  from  equations 
(27)  and  (26),  respectively,  are 

*  =  -7'.3  e  =  11°26'.3 

With  0  =  21°  and  6  =  11°  26'.3  the  values  of  yw,  ya,  yc,  yg,  and 
ye,  computed  from  equations  (22),  (20),  (21),  (19),  and  (23), 
respectively,  are 

yw  =  -2.8078  ft.  yg  =  9.1226  ft. 
ya  =  4.2374  ft.  ye  =  4.5790  ft. 
ye  =  3.7721  ft. 

With  V(dxg/dty  +  (dyg/dt)z  =  26.112  f.  s.  and  <f>  =  21°,  the 
value  of  d<j>/dt,  computed  from  equations  (28)  and  (33),  is 

d$/dt  =  2.8788  radians  per  sec. 

With  d<t>/dt  =  2.8788  radians  per  sec.,  0  =  21°,  ^  =  -7'.3, 
and  6  =  11°  26'.3,  the  values  of  d^/dt,  dO/dt,  dxg/dt,  dyg/dt, 
dxa/dt,  dya/dt,  dxc/dt,  dyc/dt,  dxe/dt,  dye/dt,  and  dyw/dt, 
computed  from  equations  (41),  (42),  (28),  (33),  (29),  (34),  (30), 
(35),  (31),  (37),  and  (36),  respectively,  are  as  follows: 

d$  /dt  =     .1147  radians  per  sec.  dxc/dt  =  11.645  f.  s. 

de  /dt  =  2.9978  radians  per  sec.  dyc/dt  =    0.2708  f.  s. 

dxa/dt  =  25.709  f.  s.  dxe/dt  =  12.059  f.  s. 

dyg/dt  =  -4.5715  f.  s.  dye/dt  =  -2.4401  f.  s. 

dxa/dt  =  11.645  f.  s.  dyw/dt  =    4.741  f.  s. 
dya/dt  =    0.2708  f .  s. 

Substituting  in  equation  (13)  the  values  of  the  coordinates 
and  the  velocities  just  determined,  together  with  the  values  of 
the  coordinates  when  the  gun  is  from  battery  and  the  values  of 
the  masses  and  moments  of  inertia  from  Table  3,  and  solving  for 
R,  we  obtain 

R  =  37312  Ibs. 

56.  Determination  of  the  Values  of  the  Forces  from  Equations 
(1)  to  (12),  Inclusive.  Gun  Fired  at  0°  Elevation.  —  The  first 
step  in  the  solution  of  equations  (1)  to  (12),  inclusive,  is  to  de- 
termine the  values  of  all  the  accelerations  in  terms  of  d^^/dt2. 
To  do  this  it  is  again  necessary  to  obtain  the  values  of  dfy/dt, 


DETERMINATION  OF   THE  FORCES  87 

d^/dt,  and  dd/dt  this  time,  however,  corresponding  to  the 
position  of  the  parts  at  the  instant  of  maximum  pressure  of  the 
powder  gases.  As  stated  in  article  44,  page  71,  it  will  be  sufficient 
to  consider  that  the  values  of  the  coordinates  and  the  angles 
are  at  this  instant  the  same  as  just  before  the  gun  was  fired.  The 
value  of  V(dxg/d£)2  +  (dyg/dt)2  will  again  be  taken  as  eight- 
tenths  of  the  velocity  of  free  recoil  which  at  this  time,  as  previously 
determined,  is  10  f .  s. 


With         V(dxg/dt)2  +  (dyg/dt)2  =  .8  x  10  =  8  f  .  s. 

the  values  of  d$/dt,  dty/dt,  and  dd/dt  computed  from  equations 
(28),  (33),  (41),  and  (42),  respectively,  are 

d<f>/dt  =  .8569  radians  per  sec. 
d\f//dt  =  —  .04256  radians  per  sec. 
dd/dt  =  .9123  radians  per  sec. 

With  these  values  of  d<f>/dt,  d^/dt,  and  dd/dt  and  the  known 
values  of  the  angles,  the  values  of  all  the  accelerations  in  terms 
of  d24>/dt2  may  be  obtained  as  follows: 

From  equation  (43)  dzxa/dt2  =  9.2937  (d24>/dt2)  -  1.44  (43') 

From  equation  (44)  d2xa/dt2  =  4.2219  (d2<j>/dt2)  -  .716  (44') 

From  equation  (45)  d2xc/dt2  =  4.2219  (d2<f>/dt2)  -  .716  (45') 

From  equation  (48)  d*yg/dP  =  -.  887585  (d2<j>/dt2)  -3.741  (48') 

From  equation  (49)  d2ya/dt2  =  .098265  (d2<f>/dt2)  -  .0167  (49') 

From  equation  (50)  d*yc/dt2  =  .098265  (d2<f>/dt2)  -  .0167  (50') 

From  equation  (51)  d2yw/dt2  =  1.07299  (d2<j>/dt2)  +  3.083  (51') 
From  equation  (56)  dfy  /dt2  =  -.04421  (d2<f>/dt2)  +  .62915  (56') 

From  equation  (57)  d28  /dtz  =  1.0647  (d?<t>/dF)  -  .1083  (57') 

From  equation  (46)  d2xe/dt2  =  4.3592  (d2cj>/dt2)  -  .2363  (46') 

From  equation  (52)  d^e/dt?  =  -.30228  (d^/dt2)  -  3.408  (52') 


By  substituting  these  expressions  for  the  accelerations  and  the 
value  of  R  =  37312  Ibs.  in  equations  (1)  to  (12),  inclusive,  the 
number  of  unknown  quantities  therein  will  be  reduced  to  twelve 
and  the  equations  can  be  readily  solved. 

Rewriting  the  equations  with  these  substitutions  and  replac- 
ing the  symbols  of  the  constants  therein  by  their  values  from 


88  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

Table  3  we  have,  since  the  value  of  ^  in  the  case  under  consider- 
ation is  0  and  the  value  of  F  corresponding  to  a  powder  pressure 
of  36000  Ibs.  per  sq.  in.  is 

36000  x  TT  (6)2/4  =  1017860  Ibs., 

1017860  -  P  -  P2  =  396.88  [9.2937  (d*<j>/dt2)  -  1.44]  (!') 

-12764  -  P!  +  P3  =  396.88  [-.887585  (d2<j>/dt2)  -  3.741]  (2') 

-5.5P3  =  14967  [-.04421  (d^/dt2)  +  .62915]  (3') 

P  -  P5  +  P6  =  187.84  [4.2219  (d^/dt2}  -  .716]  (4') 

P!  +  P4  -  6040.8  -  .15  P6  -  37312  -  19797 


-  615.58  [1.07299  (d2<f>/dt2)  +  3.083] 
=  187.84  [.098265  (ffij/dP)  -.0167] 

5.0718  P  -  .9878  P!  -  4.2224  P6  -  .975  [.15  P 
+  37312  +  19797  +  615.58  {  1.07299  ( 


(5') 


(6') 


+  3.083|]  =  1250.4  ( 

P5  =  .02327  P7  -  .02327  P8  -  .005  X  .99972  P7  ._, 

-  .005  x  .99972  P8  =  78.61  [4.2219  (d^/dt2)  -  .716]] 

.99972  P7  +  .99972  P8  -  P4  -  .005  X  .0237  P7 
-.005  x  .0237  P8  -2528  (8') 

=  78.61  [.098265  (d2<f>/dt2)  -  .0167] 

.4653  P5  -  .6651  P4  +  2.4875  P7  -  1.6792  P8  +  .4255 

x  .005  P7  +.4255  x  .005  P8  =  0        (9') 

P2  +  P9  =  21.984  [4.3592  (d2<t>/dt2}  -  .2363  (10') 

Pio  -  P3  -  707  =  21.984  [-.30228  (d2<j>/dtz)  -  3.408]     (11') 

4.6347  P2  +  .3214  P3  -  4.0944  P9  +  .2839  P10 

=  155.08  [1.0647  (d^/dt2)  -  .1083]         (12') 

SOLUTION  OF  EQUATIONS   (!')   TO   (13').  INCLUSIVE. 

From  equation  (3')  P3  =  120.3  (d2<j>/dt2)  -  1712  (3") 

Substituting  this  value  of  P3  in  equation  (2') 

Pj  =  472.57  (d2<f>/dt2)  -  12991.3  (2") 

Substituting  the  value  of  P3  in  equation  (II7) 

P10  =  113.66  (d^/d?}  -  1079.92  (11") 


DETERMINATION  OF  THE  FORCES  89 

Substituting  in  equation  (12')  the  values  of  P3  from  equation 
(3"),  Pio  from  equation  (11"),  and  P9  from  equation  (10'), 

P2  =  55.739  (cP0/(ft2)  +  93.80  (12") 

From  equation  (!')  P  =  1018337.7  -  3744.24  (d2<j>/dt2)  (1") 

From  equation  (6')  P6  =  1171851  -  4888.27  (d^/dP)  (6") 

From  equation  (4')  P5  =  2190323.19  -  9425.41  (d^/dtz)  (4") 

From  equation  (5')  P4  =  253813.38  -  26.828  (d^/dtz)  (5") 

In  equation  (7')  the  coefficient  of  P7  and  P8  when  the  two 
terms  containing  P7  and  the  two  containing  P8  are  consolidated  is 
.02827,  and  in  equation  (8')  when  these  terms  are  consolidated 
the  coefficient  is  .9996.  Therefore,  by  multiplying  equation  (7') 

9996 
by  '          and  adding  the  result  to  equation  (8'),  P?  and  P8  are 


eliminate^  and 

35.359  P5  -  P4  -  2528  =  35.359  x  78.61  [4.2219  (d*<t>/dt2) 

-  .716]  +  78  61  [.098265  (d2<t>/dt2)  -  .0167]  (7") 

Substituting  for  P5  and  P4  in  equation  (7")  their  values  from 
equations  (4")  and  (5") 

344493  (d*<t>/dt2)  =  77193650  (7'") 

Whence         <P<j>/dP  =  224.08  radians  per  sec.  per  sec.  (7/F) 

With  the  value  of  d*<t>/dtz         =  224.08  we  obtain 

From  equation  (1")     P  =  179338  Ibs.  (!'") 

From  equation  (2")    P!  =    92902  Ibs.  (2"') 

From  equation  (12")  P2  =    12584  Ibs.  (12"') 

From  equation  (3")    P3  =    25245  Ibs.  (3'") 

From  equation  (5")    P4  =  135748  Ibs.  (5'") 

From  equation  (4")    P5  =    78223  Ibs.  (4'") 

From  equation  (6")    P6  =    76476  Ibs.  (6'") 
/P,  =  .15  P6  =  11471  Ibs. 


From  equation   (51')   and  the  value  of  Mv  =  615.58  from 
Table  3, 

Mw  (d*yw/dt2)  =  149907  Ibs.  (51") 


90  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

The  total  vertical  force  acting  on  the  lower  ends  of  the  gun 
levers  through  the  gun-lever  pins  is 

fP9  +  R  +  Wv  +  Mwd*yw/dtz  =  11471  +  37312  +  19797 

+  149907  =  218487  Ibs.         (51"') 

To  obtain  the  values  of  P7  and  P8  replace  P4  and  P5  in  equation 
(9')  by  their  values  from  equations  (5'")  and  (4"'),  respectively, 
and  solve  for  P8  in  terms  of  P7.  Then  substitute  this  value  of 
PS  and  the  numerical  values  of  P4  and  tf^/dP  in  equation  (8') 
and  solve  for  P7.  The  value  of  P8  then  follows  from  that  of  P7. 

From  equation  (9')     P8  =  1.4845  P7  -  32127  (9") 

From  equation  (8')     P7  =  69304  Ibs.  (8") 

From  equation  (9")     P8  =  70755  Ibs.  (9'") 

From  equation  (10')  after  substituting  therein  the 
values  of  P2  and  d^/dt2 

P9  =  8885  Ibs.  (10") 

From  equation  (11")  after  substituting  therein  the 
value  of  dz(f>/dt2 

Pio  =  24386  Ibs.  (11'") 

57.  Computation  of  the  Values  of  the  Forces  when  the  Gun  is 
Fired  at  Any  Angle  of  Elevation.  —  When  Fired  at  Extreme 
Angles  of  Elevation  and  Depression.  —  The  determination  of 
the  forces  acting  on  the  carriage  has  been  made  under  the  sup- 
position that  the  gun  was  fired  at  zero  degrees  elevation.  The 
method  used  is,  however,  equally  applicable  to  a  case  with  any 
other  angle  of  elevation.  In  the  latter  case  the  force  of  the 
powder  gases  will  have  a  vertical  as  well  as  a  horizontal  component, 
but  the  value  of  each  component  will  be  known  and  no  additional 
unknown  quantities  will  be  introduced  into  the  equations.  x'r 
and  y'r,  the  coordinates  of  the  axis  of  the  pin  at  the  lower  end  of 
the  elevating  arm,  will  vary  with  the  elevation  given  to  the  gun, 
but  their  values  corresponding  to  any  given  elevation  can  be 
readily  obtained  from  the  drawings  of  the  carriage. 

In  order  that  the  maximum  forces  on  each  part  of  the  gun 
'carriage  may  be  determined  for  use  in  so  proportioning  the 
various  parts  that  the  stresses  in  them  will  not  be  dangerously 
great,  the  forces  should  be  computed  not  only  for  an  elevation 


DETERMINATION  OF  THE  FORCES  91 

of  the  gun  of  zero  degrees  but  also  for  the  extreme  angles  of 
elevation  and  depression  for  which  the  carriage  is  designed. 
For  the  same  reason  it  is  often  necessary,  particularly  in  the 
case  of  barbette  and  mobile  artillery  carriages,  to  compute  the 
forces  corresponding  to  different  positions  of  the  gun  in  recoil, 
generally  its  positions  in  and  from  battery. 

58.  Maximum  Values  of  the  Forces  on  a  Disappearing  Carriage 
during  Recoil.  —  Velocity  and  Acceleration  Curves  of  the  Re- 
coiling Parts.  —  In  the  case  of  a  disappearing  carriage  of  the 
service  type  the  peculiar  constraint  of  the  moving  parts  during 
recoil  is  such  that  the  maximum  accelerations  of  some  of  the 
parts  occur  somewhat  later  than  the  instant  of  maximum  powder 
pressure.  However,  by  plotting  the  velocity  curves  of  the  vari- 
ous parts  from  the  relations  established  in  equations  (14)  to 
(42),  inclusive,  it  is  seen  that  the  maximum  accelerations  of  all 
parts  occur  at  times  sufficiently  near  that  of  the  maximum  powder 
pressure  to  warrant  the  assumption  that  the  values  of  the  forces 
computed  for  that  instant  vary  but  little  from  their  maximum 
values. 

In  plotting  the  velocity  curves  referred  to,  the  velocity  of  recoil 
of  the  top  carriage  is  first  measured  by  a  Sebert  velocimeter  and 
may  then  be  plotted  both  as  a  function  of  time  and  of  space. 
For  any  measured  velocity  the  corresponding  coordinates  xc 
and  yc  of  the  center  of  mass  of  the  top  carriage  are  known  and 
from  them  may  be  obtained  by  either  of  equations  (16)  or  (21) 
the  corresponding  value  of  <£.  The  corresponding  values  of  the 
coordinates  of  the  centers  of  mass  of  the  other  moving  parts, 
and  the  values  of  the  angles  ^  and  6,  can  then  be  obtained  from 
equations  (14)  to  (27),  inclusive.  The  corresponding  velocities, 
linear  and  angular,  of  the  other  moving  parts  can  now  be  de- 
termined from  the  measured  velocity  of  the  top  carriage  and 
equations  (28)  to  (42),  inclusive.  As  the  velocity  of  the  top 
carriage  is  measured  both  as  a  function  of  time  and  of  space,  the 
velocity  curves  of  the  other  moving  parts  may  also  be  plotted 
as  functions  either  of  time  or  of  space.  The  accelerations  are 
obtained  by  measuring  the  angles  which  the  tangents  to  the 
velocity  curves  as  a  function  of  time  make  with  the  axis  of  time. 
The  tangents  of  these  angles  give  the  required  accelerations, 
which  may  then  be  plotted  either  as  functions  of  time  or  of  space. 


92  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

Having  plotted  an  acceleration  curve  of  one  moving  part,  the 
acceleration  curves  of  the  other  moving  parts  may,  if  desired,  be 
plotted  from  the  data  given  by  the  plotted  acceleration  curve 
and  its  corresponding  velocity  curves  as  a  function  of  time  and 
of  space,  and  the  relations  established  in  equations  (14)  to  (57), 
inclusive. 

If  it  is  desired  to  determine  the  acceleration  curves  of  the 
various  moving  parts  before  the  carriage  is  constructed,  it  may 
be  done  in  the  manner  to  be  described  later  in  article  66. 

59.  Computation  of  the  Values  of  the  Forces  at  Any  Instant 
While  the  Projectile  is  in  the  Bore.  —  If  we  wish  to  compute 
the  values  of  the  forces  on  the  parts  of  the  carriage  at  any  time 
while  the  projectile  is  in  the  bore  other  than  the  time  of  maximum 
powder  pressure,  the  method  to  be  followed  is  similar  to  that 
described  for  the  instant  of  maximum  powder  pressure.  For 
example,  suppose  it  is  desired  to  compute  the  values  of  the  forces 
at  the  instant  the  projectile  leaves  the  bore.  The  value  of  R, 
the  constant  resistance  of  the  recoil  cylinder,  will  be  equal  to  37312 
Ibs.  as  before.  The  value  of  the  force  F  at  this  instant  can  be 
obtained  from  the  curve  of  pressures  while  the  projectile  is  in 
the  bore  plotted  by  the  methods  of  interior  ballistics.  The 
velocity  of  the  gun  in  restrained  recoil  can  be  determined  very 
closely  as  before  from  a  consideration  of  the  data  given  by  Table 
2,  and  the  value  of  <t>  may  be  obtained  from  those  data  and  equa- 
tion (14).  The  values  of  the  angles  ^  and  6  and  of  the  coordinates 
of  the  centers  of  mass  at  the  instant  the  projectile  leaves  the  bore 
can  then  be  obtained  from  equations  (14)  to  (27),  inclusive.  The 
value  of  d<j>/dt  can  be  found  from  the  velocity  of  the  gun  in  re- 
strained recoil  selected  after  consideration  of  Table  2,  and  equa- 
tions (28)  and  (33).  The  values  of  d^/dt  and  de/dt  then  follow 
from  equations  (41)  and  (42),  respectively,  and  the  values  of  the 
accelerations  in  terms  of  d2<j>/dt2  from  equations  (43)  to  (57), 
inclusive. 

By  substituting  these  data  together  with  the  values  of  the 

constants  from  Table  3  in  equations  (1)  to  (12),  inclusive,  they 

j% , 

may  be  readily  solved  as  before  giving  the  value  of  -^  and  the 

values  of  the  forces  P  to  PIO  at  the  instant  the  projectile  leaves 
the  bore. 


DETERMINATION  OF  THE  FORCES  93 

60.  Computation  of  the  Values  of  the  Forces  at  any  Instant 
after  the  Powder  Gases  have  Ceased  to  Act  on  the  Gun.  —  After 
the  powder  gases  have  ceased  to  act  on  the  gun  the  angular  veloc- 
ity d<t>/dt  corresponding  to  any  assumed  value  of  0  can  be  ob- 
tained from  equation  (13),  or  more  readily  from  equation  (13") 
as  explained  later  in  article  64.     The  values  of  \(/,  6,  d^/dt,  dQ/dt, 
of  the  coordinates  of  the  centers  of  mass;  and  of  the  accelerations 
in  terms  of  d2(j>/dt2,  can  then  be  determined  in  the  manner  already 
explained.     The  value  of  R  is  37312  Ibs.  as  before. 

Substituting  these  data  and  the  values  of  the  constants  from 
Table  3  in  equations  (1)  to  (12),  inclusive,  and  making  F  =  0 
since  the  powder  gases  have  ceased  to  act  on  the  gun,  the  equa- 
tions may  be  solved  giving  the  values  of  d~^/dt-  and  the  forces 
P  to  Pio  corresponding  to  the  assumed  value  of  <£.  A  negative 
value  for  d^/dP  obtained  in  the  solution  of  equations  (1)  to  (12) 
indicates  that  the  rotation  of  the  gun  levers  is  being  retarded. 

61.  Centripetal  Accelerations.  —  It  will  be  noted  that  each 
expression  for  the  value  of  the  different  accelerations  in  terms  of 
the  angular  acceleration  d~$/dP,  equations  (43)  to  (57),  inclusive, 
contains  at  least  one  term  in  which  the  square  of  an  angular 
velocity  occurs.     This  is  due  to  the  rotary  motion  of  the  parts 
and    the  consequent  centripetal    accelerations  along    the  radii 
equal  to  V2/p  =  pu2  in  which  p  is  the  radial  distance  of  the 
center  of  mass  of  the  body  from  the  center  about  which  it  is  rotat- 
ing, V  is  the  linear  velocity  of  the  center  of  mass,  and  o>  the  angular 
velocity  of  the  rotating  body.     The  components  of  the  centripetal 
accelerations  in  the  directions  of  the  axes  of  X  and  Y  must  enter 
in  the  expressions  for  the  various  linear  accelerations  in  those 
directions,  and  due  to  the  relations  between  the  variable  angles  the 
expressions  for  the  angular  accelerations  d?-Q/<W-  and  d-\l//dt2  will 
also  contain  terms  involving  the  squares  of  the  angular  velocities. 

As  a  general  rule,  however,  when  forces  whose  duration  is  very 
short,  like  those  of  the  powder  gases  in  a  gun,  are  considered,  the 
accelerations  produced  by  them  will  be  relatively  very  great  as 
compared  with  the  velocities  because  of  the  very  short  period 
of  time  during  which  the  forces  act.  Referring  to  the  expres- 
sions for  the  various  accelerations  given  on  pages  87  and  88,  the 
numerical  terms  represent  the  values  of  the  terms  into  which  the 
squares  of  the  angular  velocities  entered;  and  as  the  angular 


94  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

acceleration  d-<t>/dP  has  been  determined  to  be  224.08  radians 
per  sec.  per  sec.,  it  follows  that  the  terms  containing  the  squares 
of  the  angular  velocities  are  of  relatively  little  importance  as 
regards  their  effect  on  the  computed  values  of  the  forces.  On 
this  account  and  because  of  the  labor  involved  in  considering 
them,  it  is  customary  to  neglect  such  terms  in  computing  the 
forces  brought  upon  a  gun  carriage  when  the  gun  is  fired.  Had 
this  been  done  in  the  case  under  consideration,  it  would  not  have 
been  necessary  to  determine  any  of  the  angular  velocities  of  the 
parts  at  the  instant  of  maximum  powder  pressure,  and  the  last 
terms  of  the  acceleration  equations  on  pages  87  and  88  would 
not  appear. 

62.  Effect  of  the  Movement  of  the  Parts  on  the  Intensities  of 
the  Forces.  —  It  is  well  to  consider  the  effect  of  permitting  move- 
ment of  the  parts  of  the  gun  carriage  when  the  gun  is  fired.  As  a 
result  of  such  movement  the  total  force  brought  upon  the  upper 
ends  of  the  gun  levers  is  but  201970  Ibs.,  this  being  the  square 
root  of  the  sum  of  the  squares  of  its  horizontal  and  vertical  com- 
ponents P  and  PI.  The  total  force  on  the  upper  end  of  the  ele- 
vating arm  is  28208  Ibs.,  its  horizontal  and  vertical  components 
being  P2  and  P3,  respectively.  Were  the  gun  levers  rigidly  fixed 
in  position  the  whole  force  of  the  powder  gases,  1017860  Ibs., 
would  be  brought  against  their  upper  ends;  but  when  the  parts 
of  the  carriage  are  permitted  to  move,  as  in  this  case,  by  far  the 
larger  part  of  this  force  is  absorbed  in  giving  acceleration  to  the 
gun  without  causing  any  stress  in  the  carriage  parts. 

If  it  were  not  for  the  translation  of  the  center  of  mass  of  the 
gun  levers  to  the  rear  on  the  top  carriage  during  recoil  of  the  gun, 
the  gun  and  counterweight  would  have  about  the  same  acceler- 
ation. Comparing  the  mass  of  the  counterweight,  615.58,  with 
that  of  the  gun,  396.88,  it  will  be  seen  that  if  the  top  carriage 
were  not  permitted  to  move  to  the  rear,  more  than  half  of  the 
force  exerted  by  the  powder  gases  would  have  to  be  transmitted 
through  the  gun  levers  to  the  counterweight  in  order  to  give  it 
about  the  acceleration  of  the  gun.  The  design  of  the  carriage  in 
this  case  would  of  course  have  to  be  changed  for  the  counterweight 
would  have  to  move  in  the  arc  of  a  circle  instead  of  in  a  vertical 
line  as  at  present.  The  construction  of  the  carriage,  which  per- 
mits the  center  of  mass  of  the  gun  levers  to  translate  to  the  rear 


DETERMINATION  OF  THE  FORCES  95 

on  the  top  carriage,  materially  reduces  the  acceleration  of  the 
counterweight  and  permits  the  use  of  much  lighter  gun  levers 
than  would  otherwise  be  possible.  The  acceleration  of  the  gun 
in  the  horizontal  direction  at  the  instant  of  maximum  powder 
pressure  obtained  from  equation  (43')  by  making  d?<j>/dt2  = 
224.08  radians  per  sec.  per  sec.  is  2078.42  ft.  per  sec.  per  sec.,  and 
its  acceleration  in  the  vertical  direction  obtained  from  equation 
(48')  is  202.37  ft.  per  sec.  per  sec.  The  acceleration  of  the  counter- 
weight occurs  only  in  the  vertical  direction  and  its  value  at  the 
instant  of  maximum  powder  pressure  obtained  from  equation 
(51')  is  only  243.21  ft.  per  sec.  per  sec.,  or  about  one-ninth  of  the 
horizontal  acceleration  of  the  gun. 


DETERMINATION  OF  THE  PROFILE  OF  THE  THROTTLING 
GROOVES  IN  THE  RECOIL  CYLINDER. 

63.  Formula  for  the  Area  of  Orifice.  —  Experimental  Modi- 
fication to  Allow  for  the  Contraction  of  the  Liquid  Vein.  —  The 

area  of  orifice  is  given  by  equation  (20),  page  287,  Lissak's  Ord- 
nance and  Gunnery,  which  is  as  follows: 

a2  =  7AW/2  gP  (58) 

in  which  a  is  the  area  of  orifice  in  square  feet,  A  is  the  effective 
area  of  the  piston  in  square  feet,  vr  is  the  velocity  of  restrained 
recoil,  P  is  the  total  pressure  on  the  piston,  g  is  the  acceleration 
due  to  gravity,  and  7  is  the  weight  of  a  cubic  foot  of  the  liquid  in 
the  recoil  cylinder. 

This  expression,  however,  does  not  take  into  account  the  con- 
traction of  the  liquid  vein.  From  actual  measurements  of 
velocities  of  recoil  and  the  corresponding  pressures  in  the  recoil 
cylinders,  made  at  the  Sandy  Hook  Proving  Ground  with  a 
Sebert  velocimeter  and  a  special  form  of  the  ordinary  indicator 
used  for  indicating  the  steam  pressure  corresponding  to  different 
positions  of  the  piston  in  the  cylinder  of  a  steam-engine,  it  has 
been  determined  that  the  value  of  the  area  of  orifice  required  to 
give  the  desired  pressure  on  the  piston  is  obtained  by  substituting 
for  the  velocity  of  flow  of  the  liquid  through  the  orifice 

»i  =  vrA/a  (59) 


96  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

given  by  equation  (16),  page  287,  Lissak's  Ordnance  and  Gunnery, 

a  larger  value 

vic  =  bvi  +  c  =  Q)vrA/a)  +  c  (60) 

in  which  6  and  c  are  experimental  constants  that  vary  with  the 
diameter  of  the  recoil  cylinder,  the  velocity  of  recoil,  and  the 
shape  of  the  orifice. 

Following  the  lines  of  the  discussion  on  page  287,  Lissak's 
Ordnance  and  Gunnery,  the  pressure  required  to  produce  a  ve- 
locity of  flow  Vic  through  an  orifice  is  the  pressure  due  to  a  column 
of  liquid  whose  height  is  given  by  the  equation 

v*u  =  2gh  (61) 

Substituting  for  vtc  its  value  from  equation  (60)  and  solving 

for  h,  we  have 

h  =  l(bvrA/a)+cY-/2g  (62) 

The  weight  of  a  cubic  foot  of  the  liquid  being  7,  the  weight  of 
a  column  whose  area  of  cross-section  is  equal  to  that  of  the  piston 
is  Ayh.  Ayh  is,  therefore,  P,  the  pressure  on  the  piston,  and 
multiplying  both  sides  of  equation  (62)  by  Ay  we  obtain 

p  =  Ayh  =  [(bVrA/a)  +  c]2Ay/2g  (63) 

Solving  equation  (63)  for  a  we  have 

a  =  bVrA/[V2  gP/Ay  -  c]  (64) 

which  is  the  form  of  equation  used  in  place  of  equation  (58) 
when  it  is  desired  to  take  into  account  the  contraction  of  the 
liquid  vein.  The  values  of  b  and  c  for  the  6-inch  disappearing 
carriage,  model  of  1905  Ml,  have  been  taken  as  1.3  and  122, 
respectively.  Making  these  substitutions  in  equation  (64)  it 
becomes 

a  =  1.3  vrA/[ V2  gP/Ay  -  122]  (65) 

To  express  the  area  of  orifice  and  the  area  of  the  piston  in 

square  inches  divide  a  and  A  by  144  and  \/A  by  V144  obtaining 

a°"  =  1.3  0rAn"/[\/288  gP/Au"  y  -  122]  (66) 

The  density  of  oil  used  in  the  cylinder  of  the  carriage  being  .85, 
the  weight  of  a  cubic  foot  of  the  liquid  is 

7  =  .85  x  62.5  =  53.125  Ibs. 

The  total  resistance  of  the  recoil  cylinder  is  37312  Ibs.  of 
which  about  1100  Ibs.  is  the  friction  in  the  stuffing-box  against 


DETERMINATION  Of  THE  FORCES  97 

the  piston-rod.     The  total  pressure  of  the  liquid  against  the  piston 

is,  therefore, 

P  =  37312  -  1100  =  36212  Ibs. 

The  diameter  of  the  piston  is  7.23  ins.  and  that  of  the  piston- 
rod  is  3.25  ins.,  so  that  the  effective  area  of  the  piston  is 

A°"  =  TT  K7.23)2  -  (3.25)2J/4  =  32.76  sq.  ins. 

Substituting  these  values  of  7,  P,  and  A  in  equation  (66)  and  tak- 
ing g  =  32.2,  we  have 

r-iii  -L.O    X    O^.lOVr  if\ir\  //»rr\ 

an    =  f  288  x  32.2  X36212U— ^  =  '^  "'          (6?) 
(     32.76  x  53.125     ) 

The  piston-rod  of  this  carriage  is  stationary,  and  since  the  recoil 
cylinder  is  attached  to  the  cross-head  to  which  the  counterweight 
is  also  attached,  the  cylinder  has  the  same  movement  as  the 
counterweight,  and  we  may  write 

an"  =  .1342  dyw/dt  (68) 

64.  Profile  of  the  Throttling  Grooves.  —  The  areas  of  the 
throttling  grooves  are  calculated  to  correspond  to  the  setting  of 
the  throttling  valve-stem  in  the  4th  notch  which  gives  an  opening 
through  the  valve  of  .17  sq.  in.  In  addition  the  piston  has  a 
clearance  in  the  cylinder  of  .02  in.  on  the  diameter  which  gives 
a  constant  area  for  the  escape  of  the  liquid  around  the  piston 
equal  to 

TT  [(7.25/2)2   -  (7.23/2)2]  =  .2275  sq.  in. 

Both  of  these  areas  must  be  subtracted  from  an"  to  get  the 
area  of  the  throttling  grooves.  There  are  two  such  grooves  and 
they  have  a  uniform  width  of  1.25  ins.,  so  that  their  depth  d 
at  any  point  corresponding  to  a  velocity  dyw/dt  of  the  counter- 
weight will  be 

,      aa//  -  .17  -  .2275  _  .1342  (dyw/df)  -  .3975 

2  x  1.25  2.50 

or  d  =  .05369  (dyw/dt}  -  .159  (69) 

The  profile  of  each  throttling  groove  will  be  a  curve  whose 
ordinates  are  the  values  of  d  given  by  equation  (69)  for  various 
values  of  dyw/dt,  and  whose  abscissas  are  the  values  of  yw  corre- 
sponding to  the  same  values  of  dyw/dt. 


98  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

The  value  of  yw  for  any  assumed  value  of  0  is  given  by  equation 
(22).  The  value  of  d<t>/dt  for  any  assumed  value  of  <£  equal  to 
or  greater  than  21°,  the  value  of  <£  corresponding  to  the  instant 
when  the  powder  gases  cease  to  act  on  the  gun,  can  be  obtained 
from  equation  (13),  and  with  this  value  of  d$/dt  the  correspond- 
ing value  of  dyw/dt  can  be  obtained  from  equation  (36).  Sub- 
stituting in  equation  (13)  the  values  of  the  linear  and  angular 
velocities  in  terms  of  d^/dt  from  equations  (28)  to  (42),  inclusive, 
solving  for  d<j>/dt,  and  representing,  for  the  sake  of  simplicity  of 
expression,  the  coefficients  of  d$/dt  in  equations  (41)  and  (42) 
by  X  and  U,  respectively,  we  obtain 


[R  (y"»  -  tt.)  +  Ww  (y"w  -  yw)  +  Wa  (y"a  -  ya 
+WC  (y"c  -  yc)  -  Wg  (yg  -  y"g]  -  We(ye  - 


dt 


(13') 


\  Mg  [a  cos  <j>  +  c  cos  (<t>  —  /3)]2  +  \  M  [a  tan  a  cos 
-  c  sin _(0  -  /3)]2  +  £  2mr2BX2  +  %  Ma[a  cos  <£]2 
+  £Ma  [a  tan  a  cos  <£]2  +  £  2mr2a  +  \  Mc  [a  cos  </>p 
+  i  Mc  [a  tan  a  cos  0]2  +  \  Me  [I  cos  &  U]2 
+  %Me[l  sin  0C7]2  +  \  2mr2eU2 
+  I  Mu,  [a  tan  a  cos  0  +  a  sin  0]2 

in  which 

„.  _sin0  [acos</>  +  ccos(<^>  —  ff)]  +cosg  [a  tan  a  cos  0—  csin(0  —  /3)] 
d  [sin  0  sin  ^  +  cos  6  cos  i/'] 

TT      a  cos  <j>  +  c  cos  (<f>  —  (3)  —  d  sin  ^T 
and  U  =  -  — r^ 

ocos  0 

Substituting  in  equation  (13')  the  known  values  of  the  resistance 
R,  of  the  masses,  moments  of  inertia,  and  the  ordinates  of  the 
centers  of  mass  when  the  parts  are  in  the  recoiled  position,  it 
becomes 


[333719  -  57109  yv  -  6041  ya  -  2528  ye  -  12764  yg 


dt 


198.41  [a  cos  0+c  cos  (0  -/3)]2+ 198.44  [a  tan  a  cos  <f> 
-c sin  (0  -  /3)]2  +  7483.5 X2  +  93.92  [a  cos  <f>}2 
+  93.92  [a  tan  a  cos  tf>]2  +  625.2  +39.305  [a  cos  0]2 
+  39.305  [a  tan  «  cos  <£]2  +  10.992  [I  cos  0C7]2 
+10.992  [I  sin  0C7]2  +  77.54  U2 
+  307.79  [a  tan  a  cos  0+  a  sin  <f>]2 


DETERMINATION  OF  THE  FORCES 


99 


The  first  step  in  the  solution  of  equation  (13")  to  determine 
the  value  of  d<j>/dt  corresponding  to  any  assumed  value  of  <£, 
is  to  calculate  ^  and  0  from  equations  (27)  and  (26),  respectively, 
and  from  them  and  the  assumed  value  of  <£,  the  corresponding 
values  of  X  and  U.  With  0,  ^,  0,  X,  and  U  known,  the  value  of 
the  denominator  of  equation  (13")  may  be  determined.  The 
values  of  yw,  ya,  yc,  yg,  and  ye  may  then  be  obtained  from  equations 
(22),  (20),  (21),  (19),  and  (23),  respectively,  and  the  value  of  the 
numerator  of  equation  (13")  determined.  The  value  of  d<j>/dt 
follows.  In  this  manner  the  values  of  d$/dt  for  $  =  30°,  40°, 
50°,  60°,  70°,  80°,  and  85°  have  been  calculated.  These  values, 
together  with  the  value  of  d$/dt  for  21°  already  determined,  and 
the  corresponding  values  of  \{/,  6,  yw*  dyw/dt,  and  d,  equation 
(69),  are  shown  in  the  following  table,  in  which  are  also  shown 
for  comparison  with  dyw/dt  the  corresponding  velocities  vc  of 
the  top  carriage  determined  by  obtaining  dxa/dt  from  equation 
(29)  and  dividing  it  by  cos  a. 

TABLE  4. 


d<t>/dt 

dyjdt 

d 

"a     ' 

* 

* 

e 

inches. 

radians 

(O 

inches. 

ft.  per  sec. 

per  sec. 

ft.  per  sec. 

21° 

-0°  7'.3 

11°  26'.3 

2.31 

2.8788 

4.741 

0.096 

11.645 

30° 

+0°  43'.8 

20°  41'.5 

5.97 

2.8295 

6.378 

0.183 

10.621 

40° 

+2°  37'.5 

30°  38'.8 

11.34 

2.7213 

7.790 

0.259 

9.036 

50° 

+5°4'.2 

40°  6'.7 

17.90 

2.5915 

8.770 

0.312 

7.220 

60° 

+7°  25'.5 

48°  50'.7 

25.44 

2.3927 

9.100 

0.330 

5.126 

70° 

+8°  50'.5 

56°  30'.3 

33.75 

2.0592 

8.456 

0.295 

3.053 

80° 

+8°  6'.7 

62°  34' 

42.56 

1.4214 

6.091 

0.168 

1.070 

85° 

+6°  34'.4 

64°  40' 

47.07 

0.7326 

3.1692 

0.111 

0.2767 

88° 

+5° 

65°  31'.4 

49.79 

0 

0 

0 

0 

As  the  profile  of  the  throttling  groove  is  a  smooth  curve  the 
number  of  points  given  in  Table  4  will  ordinarily  be  sufficient 
to  enable  it  to  be  plotted  from  the  abscissa  yw  =  2.31  ins.  to  the 


*  For  convenience  in  discussing  the  profile  of  the  throttling  groove,  yw  will 
hereafter  be  referred  to  the  center  of  mass  of  the  counterweight  when  the  gun 
is  in  battery  instead  of  to  the  center  of  coordinates  heretofore  assumed.  It 
will  also  be  expressed  in  inches.  Its  values  under  these  conditions  are  obtained 
by  adding  the  value  of  h  =  3  ft.  to  the  values  of  yw  obtained  from  equation  (22) 
and  multiplying  the  results  by  twelve  to  reduce  feet  to  inches. 


100  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

end,  although  after  plotting  it  from  the  table  it  may  be  found 
desirable  to  calculate  one  or  two  more  points  on  the  curve  to 
determine  with  certainty  the  location  of  its  maximum  ordinate 
and  the  point  at  which  it  ends.  On  account  of  the  subtractive 
term  in  equation  (69)  the  throttling  groove  does  not  commence 
at  the  point  where  yw  is  zero  and  does  not  extend  to  the  point 
where  yw  is  49.79  ins.  From  equation  (69),  d  will  be  equal  to 
zero  when  dyw/dt  is  equal  to  .159/.05369  =  2.961  f.  s.  Until 
dyw/dt  attains  this  value  the  groove  will  not  commence  and  it 
will  cease  as  soon  as  dyw/dt  decreases  to  this  value  after  having 
reached  its  maximum.  The  point  where  the  groove  ends  becomes 
apparent  at  once  when  its  profile  is  plotted  to  a  large  scale  with  a 
sufficient  number  of  calculated  points.  The  point  where  it  com- 
mences is  determined  as  described  in  the  next  paragraph. 

Equations  (13),  (13'),  and  (13")  are  not  applicable  to  values  of 
<f>  less  than  21°  for  then  the  powder  gases  are  still  acting  on  the 
gun.  d$/dl  can  not,  therefore,  be  obtained  from  equation  (13") 
for  values  of  yw  less  than  2.31  ins.  Owing  to  the  rapidly  changing  'y 
force  exerted  by  the  powder  gases  on  the  gun  and  to  the  imprac- 
ticability of  deducing  a  simple  equation  expressing  the  value  of  £ 
the  force  as  a  function  of  time  or  space,  it  is  not  practicable  to 
obtain  by  analytical  methods  reliable  values  for  dyw/dt  while  ,  j 
the  powder  gases  are  acting,  although  these  values  may  be  ob- 
tained approximately  from  the  curve  of  free  recoil  of  the  gun  as  a 
function  of  space.  In  the  case  of  those  barbette  and  mobile 
artillery  carriages  in  which  the  resistance  to  recoil  is  constant  and 
applied  in  the  direction  of  the  axis  of  the  gun,  the  curve  showing 
the  velocity  of  restrained  recoil  may  be  accurately  determined 
by  graphical  methods  as  described  in  Par.  166,  pages  283  and  284, 
Lissak's  Ordnance  and  Gunnery.  The  general  form  of  the  profile 
of  the  throttling  grooves  for  such  carriages  is  well  known  and  it 
is  of  the  same  character  beyond  the  point  where  the  powder  gases 
cease  to  act  as  the  profile  of  the  throttling  groove  of  the  disap- 
pearing carriage.  It  is  a  reasonable  assumption,  therefore,  that 
the  part  of  the  profile  of  the  throttling  groove  which  corresponds 
to  the  position  of  the  parts  while  the  powder  gases  are  acting  is 
also  similar  in  the  two  cases.  For  values  of  yw  less  than  2.31  ins., 
therefore,  the  profile  of  the  throttling  groove  of  the  6-inch  dis- 
appearing carriage,  model  of  1905  Mi,  is  obtained  by  prolonging 


ENGINES 


DETERMINATION  OF  THE  FORCES  101 

the  curve  already  plotted  back  to  the  origin  and  giving  to  it  the 
ordinary  form. 

Each  new  type  or  model  of  carriage  is  tested  at  the  Sandy 
Hook  Proving  Ground,  and  any  necessary  changes  in  its  design 
and  in  the  areas  of  the  throttling  orifices  are  determined  there. 
These  changes  are  then  applied  to  all  carriages  of  that  model 
before  they  are  issued  to  the  service. 

Fig.  34  shows  the  complete  profile  of  the  throttling  groove 
for  this  carriage  as  it  would  be  furnished  to  the  shops  for  cutting 
the  grooves  in  the  recoil  cylinder. 

It  will  be  observed  on  examining  this  figure  that  above  the 
word  "  PISTON  "  the  curve  is  straight  for  a  distance  of  2.25  ins., 
its  ordinate  there  being  the  maximum  ordinate.  This  arrange- 
ment is  necessary  because  the  piston  is  2.25  ins.  long  and  the 
smallest  opening  between  it  and  the  wall  at  the  bottom  of  the 
groove,  or  the  controlling  orifice,  is  at  the  upper  edge  of  the  piston 
until  the  maximum  ordinate  of  the  groove  is  reached  during  recoil. 
On  account  of  the  length  of  the  cylindrical  piston  the  maximum 
orifice  could  not  be  obtained  if  the  straight  part  of  the  profile  of 
the  groove  did  not  exist,  for  before  the  upper  edge  of  the  piston 
reached  the  maximum  ordinate  the  lower  edge  would  have  passed 
it  and  the  opening  at  the  lower  edge  would  have  become  the  con- 
trolling orifice.  The  straight  part  of  the  profile  permits  the  ori- 
fice at  the  upper  edge  of  the  piston  to  be  the  controlling  orifice 
until  it  reaches  the  maximum  ordinate,  after  which  the  orifice 
at  the  lower  edge  of  the  piston  becomes  the  controlling  ori- 
fice until  recoil  ends. 

65.  Velocities  of  the  Counterweight  and  the  Top  Carriage 
Compared.  —  Advantages  of  the  Recoil  Cylinder  in  the  Counter- 
weight and  of  the  Counter-Recoil  Buffer  Acting  Against  the  Top 
Carriage.  —  The  velocities  of  the  top  carriage  shown  in  the  last 
column  of  Table  4  have  been  calculated  for  comparison  with  those 
of  the  counterweight,  and  to  illustrate  by  an  example  taken  from 
a  service  carriage  the  statements  made  in  Par.  200,  page  346, 
Lissak's  Ordnance  and  Gunnery,  as  to  the  reasons  for  changing 
the  position  of  the  recoil  cylinders  of  the  disappearing  carriage 
from  the  top  carriage  to  the  counterweight  while  retaining  the 
action  of  the  counter-recoil  buffer  on  the  top  carriage.  By  ex- 
amination of  the  table  it  will  be  seen  that  the  velocity  of  the  top 


102  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

TOP  OF  PISTON    IN, 


DETERMINATION  OF  THE  FORCES  103 

carriage  toward  the  end  of  recoil  is  much  less  than  that  of  the 
counterweight,  so  that  a  recoil  cylinder  in  the  counterweight 
gives  better  control  of  the  final  movement  of  the  gun  in  recoil 
than  does  one  placed  in  the  top  carriage.  As  the  loading  angle 
and  the  height  of  the  breech  of  the  gun  are  greatly  affected  by  a 
comparatively  small  change  in  the  total  amount  of  recoil,  it  is 
important  that  this  variation  be  as  small  as  possible. 

The  relations  between  the  velocities  of  the  counterweight  and 
top  carriage  at  any  point  are  the  same  in  counter-recoil  as  in  recoil, 
so  that  when  the  gun  is  almost  in  battery  during  counter-recoil 
the  velocity  of  the  top  carriage  is  much  greater  than  that  of  the 
counterweight  and  a  counter-recoil  buffer  acting  against  the  top 
carriage  gives  better  results  than  would  one  acting  on  the  counter 
weight. 

66.  Velocity  and  Acceleration  Curves  of  the  Recoiling  Parts.  — 
Article  58  gives  a  description  of  the  method  of  plotting  the  veloc- 
ity and  acceleration  curves  of  the  recoiling  parts  of  a  disappearing 
carriage  based  upon  the  measurement  of  the  velocities  of  the  top. 
carriage  by  the  Sebert  velocimeter.  If  these  curves  are  desired 
before  the  carriage  is  manufactured  they  can  be  obtained  from  the 
velocities  of  restrained  recoil  used  in  the  calculations  of  the  profile 
of  the  throttling  grooves.  For  example,  Table  4  gives  the  velocity 
of  recoil  of  the  counterweight  as  a  function  of  space  corresponding 
to  a  number  of  positions  of  the  counterweight  in  recoil  after  the 
powder  gases  have  ceased  to  act  on  the  gun.  The  velocity  of  the 
counterweight  corresponding  to  a  number  of  its  positions  while 
the  powder  gases  are  acting  may  also  be  obtained  by  substituting 
in  equation  (69)  the  values  of  d  scaled  from  the  early  part  of  the 
profile  of  the  throttling  groove.  By  plotting  the  reciprocals  of 
the  velocities  of  the  counterweight  as  a  function  of  space  and 
integrating  under  the  curve  up  to  any  ordinate,  the  area  will 
represent  the  tune  corresponding  to  the  space  over  which  the 
counterweight  has  recoiled,  these  operations  being  the  same  as 
those  for  obtaining  the  curve  of  the  velocity  of  the  projectile  in 
the  bore  as  a  function  of  time.  Having  the  times  corresponding 
to  the  various  distances  recoiled  by  the  counterweight,  its  velocity 
of  recoil  may  be  plotted  as  a  function  of  time  as  well  as  of  space. 
The  velocity  of  recoil  of  each  of  the  other  moving  parts  and  the 
corresponding  accelerations  of  all  the  moving  parts  may  now  be 


104  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

plotted  as  a  function  either  of  time  or  space  in  a  manner  similar 
to  that  described  in  article  58. 

In  this  way  we  can  obtain,  before  the  carriage  is  built,  the  total 
time  of  recoil,  and  the  position,  velocity,  and  acceleration  of  each 
of  the  recoiling  parts  of  a  disappearing  carriage  corresponding 
to  any  interval  of  time  measured  from  the  instant  when  recoil 
began. 

67.  Method  of  Designing  a  Gun  Carriage.  —  The  forces  on 
the  various  parts  of  a  gun  carriage  having  been  determined,  it  is 
then  necessary  to  compute  the  stresses  caused  by  them  in  the 
parts  and  to  so  proportion  the  parts  that  the  stresses  will  be  kept 
within  permissible  limits.  The  method  of  doing  this  will  be  ex- 
plained in  Chapter  IV.  From  what  has  preceded  it  is  apparent 
that  the  masses,  and  therefore  the  weights,  and  the  distribution 
of  the  mass  of  each  part,  represented  by  its  moment  of  inertia, 
affect  to  a  considerable  degree  the  forces  brought  upon  the  parts 
when  the  gun  is  fired;  while  on  the  other  hand,  the  intensities  of 
the  forces  govern  in  most  cases  the  sizes  and  distribution  of  the 
mass  of  the  parts.  It  is  evident,  therefore,  that  the  design  of  a 
gun  carriage,  as  of  any  other  machine  where  the  strength  of  the 
parts  must  be  considered,  is  a  matter  of  trial  in  which  the  actual 
amount  of  work  involved  is  largely  a  matter  of  the  skill  of  the 
designer  and  his  experience  with  the  type  of  gun  carriage  or  ma- 
chine being  designed.  The  general  outline  of  the  carriage  is 
first  laid  down  on  the  drawing  and  the  dimensions  of  the  parts 
determined  approximately  by  experience  and  judgment.  The 
weights  of  the  parts  are  then  calculated  from  the  drawing  by 
obtaining  their  volumes  and  multiplying  them  by  the  weights  of 
a  unit  volume  of  the  various  materials  used.  The  centers  of 
mass  of  the  parts  and  their  moments  of  inertia  are  calculated  in 
accordance  with  the  principles  given  in  works  on  mechanics. 
(The  method  of  calculating  centers  of  mass  and  moments  of  inertia 
of  parts  from  a  drawing  will  be  illustrated  later  in  Chapter  IV.) 

The  forces  on  the  parts  are  next  calculated  and  then  the  stresses 
caused  by  them.  It  will  generally  happen  that  the  stresses  in  some 
of  the  parts  are  too  great  for  safety  and  that  in  other  parts  they 
are  smaller  than  they  need  be.  Those  parts  in  which  the  stresses 
are  too  great  must  be  increased  in  size  or  changed  in  shape  in  such 
manner  as  to  decrease  the  stresses  to  a  reasonable  figure.  If 


DETERMINATION  OF  THE  FORCES  105 

it  is  important  that  the  weight  of  the  carriage  be  kept  as  small 
as  possible,  as  is  the  case  with  all  mobile  artillery  carriages,  those 
parts  in  which  the  stresses  are  smaller  than  they  need  be  must  be 
decreased  in  size  and  weight  until  the  stresses  are  raised  to  the 
maximum  permissible  limits.  If  weight  is  not  important  and  the 
unnecessarily  large  size  of  the  parts  is  not  objectionable  otherwise, 
the  parts  may  not  be  changed. 

If  the  changes  in  the  parts  are  so  considerable  as  to  require 
a  redetermination  of  the  weights,  centers  of  mass,  moments 
of  inertia,  and  finally  of  the  forces  on  the  parts,  this  must  be  done; 
and  the  stresses  on  the  parts  due  to  the  changed  forces  must  again 
be  determined.  These  processes  are  repeated  until  the  sizes, 
weights,  and  shapes  of  the  parts  are  considered  satisfactory. 


CHAPTER  IV. 
STRESSES  IN   PARTS   OF  GUN  CARRIAGES. 

68.  Strength  of  Materials.  —  The  determination  of  the  stresses 
in  parts  of  gun  carriages  requires  a  knowledge  of  strength  of 
materials,  but  as  this  subject  is  studied  by  the  cadets  of  the  first 
class  in  the  course  of  civil  and  military  engineering  the  principles 
involved  will  not  be  deduced  here.     The  most  important  facts 
and  formulas  will,  however,  be  stated  and  briefly  explained. 

69.  Stresses  of  Tension  and  Compression.  —  Let  AB,  Fig.  35, 
be  a  rod  fixed  at  the  end  A.    If  a  force  T  acts  upon  it  in  the 
direction  of  its  axis  to  elongate  it  as  shown,  the  force  will  produce 


B 


Fig.  35. 


a  stress  of  tension  in  the  rod  which  will  be  distributed  uniformly 
over  each  cross-section  between  B  and  A,  the  intensity  of  the  stress 
per  unit  of  area  being  equal  to  the  total  force  divided  by  the  area 
of  cross-section  considered. 

If  the  direction  of  the  force  be  reversed  it  will  compress  the 
rod  producing  in  it  a  stress  of  compression.  This  stress  also  will 
be  distributed  uniformly  over  each  cross-section  between  B  and  A 
and  its  intensity  per  unit  of  area  will  equal  the  total  force  divided 
by  the  area  of  cross-section  considered. 

If  the  force  T  be  not  applied  at  the  end  of  the  rod  but  at  some 
section  nearer  A  it  will  cause  a  stress  in  the  sections  between  it 
and  A  but  none  in  those  between  it  and  the  free  end  B.  If  a 
number  of  forces  act  instead  of  one  force  T  each  will  produce  a 
stress  in  every  section  affected  by  it  and  the  algrebaic  sum  of 
these  stresses  in  any  section  will  be  the  total  stress  in  the  section. 

106 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES 


107 


.  In  general,  if  a  rod  is  in  equilibrium  under  the  action  of  a  num- 
ber of  forces,  all  those  that  are  parallel  to  its  axis  will  cause  either 
tension  or  compression  in  the  rod.  If  any  section  of  the  rod  be 
considered  the  tension  or  compression  in  it  will  be  due  to  all  of 
the  forces  acting  on  one  side  (either  side)  of  that  section;  and  if 
on  either  side  there  are  no  forces  acting  there  will  be  no  stress  in 
the  section.  All  forces  acting  toward  a  section  produce  com- 
pression therein  and  all  those  acting  away  from  it  produce  ten- 
sion therein. 

70.  Shearing  Stress.  —  If  two  flat  plates  are  laid  one  upon 
the  other  and  fastened  together  by  rivets  any  force  applied  to 
the  plates  to  slide  one  along  the  other  will  be  a  simple  shearing 
force  as  applied  to  the  rivets.  This  force  will  be  resisted  by  a 
shearing  stress  in  the  cross-sections  of  the  rivets  lying  in  the  plane 
of  contact  of  the  plates.  The  shearing  stress  will  be  distributed 
uniformly  over  the  cross-sections  of  the  rivets  and  its  intensity 
per  unit  of  area  will  be  equal  to  the  total  shearing  force  divided 
by  the  total  area  of  cross-section. 


i 
i 
i 

i 
i 
i 

M/ 


s 


B 


Fig.  36. 

If  a  solid  bar  be  considered  instead  of  the  plates  the  same  force 
would  tend  to  slide  one  part  past  the  other  in  the  same  manner 
as  if  they  were  separate  parts  fastened  together  by  rivets  as 
before,  and  the  stress  developed  between  them  would  be  one  of 
simple  shear. 

Let  AB,  Fig.  36,  be  a  block  imbedded  in  a  wall  and  S  a  force 
applied  to  the  block  through  the  center  of  the  section  next  the 
wall  as  shown.  The  force  will  tend  to  slide  that  portion  of  the 
block  projecting  from  the  wall  past  the  portion  imbedded  therein 


108  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

and  will  produce  in  the  section  next  the  wall  a  shearing  stress 
uniformly  distributed  over  it,  whose  intensity  per  unit  of  area 
will  equal  the  total  force  divided  by  the  area  of  the  section. 

If  the  force  S  be  not  applied  next  the  wall  but  at  a  distance 
from  it  as  shown  by  the  dotted  force  line  it  will  produce  a  shearing 
stress  in  every  section  between  it  and  the  wall  but  none  in  the 
sections  between  it  and  the  free  end  of  the  block.  It  will  also  pro- 
duce a  bending  stress  in  the  sections  between  it  and  the  wall 
which  will  be  discussed  later.  In  this  and  every  other  case  of  a 
shearing  stress  caused  by  a  force  that  also  produces  a  bending 
stress,  the  shearing  stress  is  not  distributed  uniformly  over  the 
section  but  varies  from  a  maximum  at  the  axis  of  the  section 
(called  the  neutral  axis)  which  is  perpendicular  to  the  force  to 
zero  at  the  points  or  lines  of  the  section  most  distant  from  the 
neutral  axis.  Thus  in  Fig.  36  aabb  is  the  section  next  the  wall. 
The  shearing  stress  therein  due  to  the  force  next  the  wall  is  uni- 
formly distributed,  but  that  due  to  the  force  shown  by  the  dotted 
line  varies  from  a  maximum  at  the  line  cc  to  zero  at  the  lines  aa 
and  bb.  It  is  ordinarily  assumed  that  a  shearing  stress,  however 
caused,  is  uniformly  distributed  over  the  section  since  when 
caused  by  a  force  that  also  produces  a  bending  stress  it  is  gener- 
ally much  smaller  than  the  latter. 

If  a  number  of  shearing  forces  act  on  a  piece,  each  will  produce 
a  shearing  stress  in  every  section  affected  by  it  and  the  algebraic 
sum  of  these  stresses  in  any  section  will  be  the  total  shearing 
stress  in  the  section. 

In  general,  if  a  piece  is  in  equilibrium  under  the  action  of  a 
number  of  forces,  all  those  that  are  perpendicular  to  its  axis  will 
cause  shear  in  the  piece,  and  the  total  shearing  stress  in  any  sec- 
tion will  be  due  to  all  the  forces  perpendicular  to  the  axis  of  the 
piece  acting  on  one  side  (either  side)  of  that  section.  If  on  either 
side  there  are  no  forces  acting  there  will  be  no  stress  in  the  sec- 
tion. 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES 
TOBSIONAL   STRESS. 


309 


71.  Rod  Fixed  at  One  End.  —  Let  AB,  Fig.  37,  be  a  rod  fixed 
at  the  end  A*  and  acted  on  at  the  free  end  by  a  force  F  perpen- 
dicular to  its  axis  as  shown.  If  this  force  intersected  the  axis 
of  the  rod  it  would  produce  in  the  rod  only  shearing  and  bending 


v 


B 


Fig.  37. 


stresses  as  already  explained,  but  having  a  lever  arm  with  respect 
to  the  axis,  it  twists  the  rod  producing  in  it  a  torsional  stress  also. 
Shearing  stresses  have  already  been  considered  and  bending  stresses 
will  be  taken  up  later,  so  that  the  twisting  or  torsional  stress  in 
the  rod  will  alone  be  discussed  at  present. 

The  torsional  moment  of  the  force  at  any  section  between  it 
and  the  fixed  end  of  the  rod  is 


Mt  =  F  x  al 


(1) 


in  which  Mt  is  the  torsional  moment,  F  is  the  intensity  of  the 
force,  and  al  is  its  lever  arm  in  inches  with  respect  to  the  axis  of 
the  rod.  The  torsional  stress  will  be  proportional  to  the  torsional 

*  In  order  that  a  rod  or  beam  may  be  fixed  at  one  or  both  ends  a  portion  of 
its  length  at  one  or  both  ends  must  be  clamped  in  some  manner  that  will  prevent 
all  movement  of  the  portions  clamped.  The  rod  may  be  clamped  by  having  its 
end  or  ends  embedded  in  a  wall  or  by  any  other  suitable  means.  When  the 
length  of  a  rod  that  is  fixed  at  one  or  both  ends  is  referred  to,  it  is  to  be  under- 
stood as  meaning  the  length  of  that  portion  which  is  not  clamped;  and  when 
the  fixed  end  of  a  rod  fixed  at  one  end  only  is  referred  to,  it  is  to  be  understood 
as  referring  to  the  section  of  the  rod  at  the  end  of  the  clamp  next  the  undamped 
portion  of  the  rod.  Similarly,  when  the  ends  of  a  rod  that  is  fixed  at  both  ends 
are  referred  to,  it  is  to  be  understood  as  meaning  the  sections  at  the  end  of 
both  clamps  next  the  undamped  portion  of  the  rod. 


110 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


moment  but  its  intensity  on  any  elementary  area  of  a  cross- 
section  will  vary  directly  with  the  distance  of  that  area  from  the 
axis.  If  a  number  of  twisting  forces  act  on  the  rod  each  will 
produce  a  torsional  moment  at  every  section  affected  by  it  and 
the  algebraic  sum  of  these  moments  at  any  section  will  be  the 
total  torsional  moment  at  the  section.  If  there  is  no  twisting 
force  at  the  free  end  of  the  rod  there  will  be  no  torsional  moment 
at  any  section  between  the  free  end  and  that  section  whose  plane 
contains  the  twisting  force  nearest  to  the  free  end. 

Rod  Fixed  at  Both  Ends.*  —  In  Fig.  37  the  rod  is  fixed  at  one 
end  only;  if  fixed  at  both  ends  as  shown  in  Fig.  38  the  force  F 
will  produce  a  torsional  moment  on  each  side  of  the  section  in 
whose  plane  it  is  contained  but  the  moment  will  not  be  equal  to 


C 


SECTION  E-E. 


Fig.  38. 


F  x  al  and  it  will  be  different  on  the  two  sides  of  the  section.  If 
c  and  Ci  represent  the  distances  of  this  section  from  the  left  and 
right  ends,  respectively,  of  the  rod,  the  torsional  moment  at  any 
section  within  the  portion  c  will  be 


Mt  =  F  x  al  x 


and  that  at  any  section  within  the  portion  Ci  will  be 
M t  =  F  x  al  x  — ^~ 


(2)t 


(3)t 


*  See  footnote,  page  109. 

t  From  Reuleaux's  Constructor. 


STRESSES  IN  PARTS  OP  GUN  CARRIAGES 


111 


the  larger  of  the  two  moments  pertaining  to  the  section  whose 
plane  contains  the  force. 

General  Case  of  Torsional  Moments.  —  When  a  rod  is  in 
equilibrium  under  the  forces  acting  on  it  and  all  the  twisting 
forces  and  couples  either  to  the  left  or  to  the  right  of  a  section 
are  known,  the  torsional  moment  at  that  section  is  the  algebraic 
sum  of  the  individual  moments  with  respect  to  the  axis  of  the  rod 
of  each  of  the  twisting  forces  and  couples  on  either  side  of  the 
section.  The  reactions  between  the  rod  and  its  supports  must 
be  included  among  the  twisting  forces  and  couples  on  either  side 
of  the  section,  and  when  this  can  be  done  it  is  unnecessary  to 
consider  whether  the  rod  is  fixed  at  one  or  both  ends.  If  on  either 
side  of  the  section  there  are  no  twisting  forces  or  couples  there  is 
no  torsional  moment  at  the  section. 

Intensity  of  Torsional  Stress.  —  Let  Ip  be  the  polar  moment 
of  inertia  of  a  cross-section  of  the  rod,  Mt  the  torsional  moment 
at  the  section,  and  S'"t  the  torsional  stress  per  unit  of  area  at  a 
unit's  distance  from  the  axis;  then,  as  shown  on  pages  38  and  39, 
Fiebeger's  Civil  Engineering, 

S'"t  =  Mt/Ip 

and  the  stress  per  unit  of  area  at  any  distance  r  from  the  axis  will 

be 

rS'"t  =  Mff/I,  (4) 

The  maximum  stress  will  evidently  occur  at  that  point  of  the 
section  most  distant  from  the  axis. 


K- X -> 


Fig.  39. 
BENDING  STRESS. 

72.   Cantilever  Subjected  to  a  Concentrated  Bending  Force.  - 

Let  AB,  Fig.  39,  be  a  rod  or  beam  fixed  at  the  end  A*  and  acted 
*  See  footnote,  page  109. 


112  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

on  at  the  free  end  by  a  force  F  whose  action  line  is  perpendicular 
to  and  intersects  the  axis  of  the  beam.  Such  a  force  has  no 
torsional  moment  at  any  cross-section  of  the  beam  but  produces 
in  each  section  a  shearing  and  a  bending  stress.  The  bending 
stress  only  will  be  considered.  The  force  F  bends  the  beam  and 
rotates  each  section  about  its  axis  perpendicular  to  the  action 
line  of  the  force,  the  amount  of  this  rotation  varying  directly  as 
the  moment  of  the  force  with  respect  to  the  section.  This  mo- 
ment is  called  the  bending  moment  of  the  force  and  is  equal  to 

M  =  Fx  (5)* 

in  which  M  is  the  bending  moment,  F  is  the  intensity  of  the  force, 
and  x  the  perpendicular  distance  in  inches  between  it  and  the 
section  under  consideration. 

The  rotation  of  the  sections  causes  the  fibres  on  top  to  be  ex- 
tended producing  in  them  a  stress  of  tension,  and  those  on  the 
bottom  to  be  compressed  producing  in  them  a  stress  of  com- 
pression. If  the  force  acted  upward  the  upper  fibres  of  the  beam 
would  be  compressed  and  the  lower  fibres  extended.  The  fibres 
intersecting  the  axes  about  which  the  sections  rotate,  called  the 
neutral  axes,  are  neither  extended  nor  compressed,  and  the  ten- 
sion or  compression  in  any  other  fibre  at  any  section  is  directly 
proportional  to  the  bending  moment  at  that  section  and  to  the 
distance  of  the  fibre  from  the  neutral  axis.  A  bending  stress, 
therefore,  is  one  in  which  the  fibres  on  one  side  of  the  neutral  axis 
are  subjected  to  a  tensile  stress  and  those  on  the  other  side  to  a 
compressive  stress. 

In  Fig.  39  the  force  acts  at  the  free  end  of  the  beam  and  every 
section  between  it  and  the  fixed  end  is  subjected  to  a  bending 
moment  and  resulting  bending  stress,  but  if  the  force  did  not  act 
at  the  free  end  there  would  be  no  bending  moment  or  bending 
stress  in  any  section  between  the  force  and  that  end.  If  a  num- 
ber of  bending  forces  act  on  the  beam,  each  will  produce  a  bending 
moment  at  every  section  affected  by  it,  and  the  algebraic  sum  of 
all  these  moments  at  any  section  will  be  the  total  bending  mo- 
ment at  the  section. 


*  The  general  expressions  for  the  bending  moments  given  by  equations  (5) 
to  (23),  inclusive,  are  taken  from  Merriman's  Text  Book  on  the  Mechanics  of 
Materials. 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES 


113 


A  beam  such  as  is  shown  in  Fig.  39  is  called  a  cantilever.  The 
maximum  bending  moment  occurs  at  the  section  at  the  fixed 
end. 

Intensity  of  Bending  Stress.  —  Let  J  be  the  moment  of  in- 
ertia of  a  cross-section  of  a  beam  with  respect  to  its  neutral  axis, 
M  the  bending  moment  at  the  section,  and  S'"  the  stress  per 
unit  of  area  at  a  unit's  distance  from  the  neutral  axis;  then,  as 
shown  on  pages  49  and  50,  Fiebeger's  Civil  Engineering,  the  stress 
per  unit  of  area  at  any  distance  y  from  the  neutral  axis  is 

S"'y  =  My /I  (6) 

Beam  Fixed  at  Both  Ends  Subjected  to  a  Concentrated  Bending 
Force.*  —  If  both  ends  of  the  beam  are  fixed  as  in  Fig.  40  the 


Fig.  40. 


force  F  will  produce  a  bending  moment  on  each  side  of  the  sec- 
tion in  whose  plane  it  is  contained.  If  I  is  the  length  of  the  beam 
in  inches  and  kl  the  distance  of  the  force  from  the  left  end,  k 
being  a  fraction  less  than  unity,  the  bending  moment  at  any 
section  on  the  left  of  the  force  at  a  distance  x  from  the  left  end  of 
the  beam  is 


M  =  -Flk  (1  -  2  k  +  A;2)  +  F  (1  -  3  k2  +  2  ks)  x 


(7) 


and  the  bending  moment  at  any  section  on  the  right  of  the  force 
at  a  distance  x  from  the  left  end  of  the  beam  is 


M  =  -Flk  (1  -  2  k  +  &2)  +  Fk  (I  -  3  kx  +  2  Wx)         (8) 
If  k  is  \,  that  is,  if  the  force  is  applied  at  the  middle  of  the  beam, 


*  See  footnote,  page  109. 


114 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


the  bending  moment  at  any  section  on  the  left  of  the  force  at  a 
distance  x  from  the  left  end  is 

TI  f          Fl  <  Fx  /r.N 

M  =--£-  +  -s-  (9) 

O  6 

and  at  any  section  on  the  right  of  the  force  at  a  distance  x  from 

the  left  end,  it  is 

Fl     F 


The  maximum  bending  moment  in  this  case  occurs  at  the  ends 
of  the  beam  and  under  the  force,  and  is 

M  =  ±Fl/S  (11) 

If  a  number  of  forces  parallel  to  F  act  on  the  beam  each  will 
cause  bending  moments  in  the  various  sections  thereof  which 
may  be  determined  from  equations  (7)  and  (8),  and  the  algebraic 
sum  of  all  those  moments  at  any  section  will  be  the  total  bending 
moment  at  the  section. 


x- 


V 


Fig.  41. 

Beam  Supported  at  Both  Ends  Subjected  to  a  Concentrated 
Bending  Force.*  —  If  the  ends  of  the  beam  are  not  fixed  but  are 
merely  supported  as  shown  in  Fig.  41,  the  bending  moment  at 
any  section  on  the  left  of  the  force  at  a  distance  x  from  the  left 
support  is 

M  =  F  (1  -  k)x  (12) 

and  at  any  section  on  the  right  of  the  force  at  a  distance  x  from 
the  left  support,  it  is 

M  =  Fk  (I  -  x)  (13) 

in  which  I  is  the  length  of  the  beam  in  inches. 

*  The  length  of  a  beam  that  is  supported  at  both  ends  is  taken  to  be  the 
distance  between  the  supports,  and  the  sections  of  the  beam  at  the  supports 
are  referred  to  as  the  ends  of  the  beam. 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES  115 

When  x  =  Qorx  =  1,    M  =  0 

and,  therefore,  the  bending  moment  is  zero  at  the  ends  of  a  beam 
that  is  merely  supported.  The  maximum  bending  moment 
occurs  under  the  force  and  is 

M  =  F  (1  -  Jfe)  Id  (14) 

If  the  force  acting  at  the  middle  of  the  beam  k  is  \  and  the  bend- 
ing moment  at  any  section  on  the  left  of  the  force  at  a  distance 
x  from  the  left  support  is 

M  =  Fx/2  (15) 

and  at  any  section  on  the  right  of  the  force  at  a  distance  x  from 
the  left  support,  it  is 

M  =  F  (I  -  3)/2  (16) 

The  maximum  bending  moment  occurs  under  the  force  and  is 

M  =  Fl/4  (17) 

If  a  number  of  forces  parallel  to  F  act  on  the  beam,  each  will 
cause  bending  moments  in  the  various  sections  thereof  which 
may  be  determined  from  equations  (12)  and  (13);  and  the  alge- 
braic sum  of  all  these  bending  moments  at  any  section  will  be 
the  total  bending  moment  at  the  section. 

Cantilever  Subjected  to  a  Uniformly  Distributed  Bending 
Force.  —  If  the  force  F  shown  in  Fig.  39  were  uniformly  dis- 
tributed over  the  length  of  the  beam  instead  of  being  concentrated 
at  the  free  end,  the  bending  moment  at  any  section  at  a  distance 
x  from  the  free  end  of  the  beam  would  be 

M  =  wx*/2  (18) 

in  which  w  is  the  intensity  of  the  force  per  linear  inch  of  the  beam. 
The  maximum  bending  moment  would  occur  at  the  fixed  end  of 
the  beam  and  would  be 

M  =  wP/2  (19) 

I  being  the  length  of  the  beam  in  inches. 

Beam  Fixed  at  Both  Ends  Subjected  to  a  Uniformly  Distributed 
Bending  Force.  —  If  the  force  in  Fig.  40  were  uniformly  dis- 
tributed over  the  length  of  the  beam  the  bending  moment  at  any 
section  would  be 

!-  (20) 


116 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


in  which  x  is  the  distance  of  the  section  from  the  left  end  of  the 
beam,  I  the  length  of  the  beam  in  inches,  and  w  the  intensity  of 
the  force  per  linear  inch  of  the  beam.  The  maximum  bending 
moment  would  occur  at  the  ends  of  the  beam  and  would  be 


M  =  -wl2/12 


(21) 


Beam  Supported  at  Both  Ends  Subjected  to  a  Uniformly  Dis- 
tributed Bending  Force.  —  If  the  force  in  Fig.  41  were  uniformly 
distributed  over  the  length  of  the  beam  between  the  supports, 
the  bending  moment  at  any  section  would  be 

wlx 


JVL    —   ~T\      ~~  — « — 


(22) 


B- 


-B 
A 


B B 


Y 

Fig.  42. 

in  which  x,  I  and  w  have  the  same  significance  as  before.  The 
bending  moment  at  the  ends  of  the  beam  would  be  zero.  The 
maximum  bending  moment  would  occur  at  the  middle  of  the 
beam  and  would  be  » 

M  =  wl*/8  (23) 

Bending  Moment  at  a  Section  of  a  Beam  Due  to  Any  Force 
Having  a  Lever  Arm  with  Respect  to  an  Axis  of  the  Section.  — 

In  the  preceding  discussion  of  bending  moments  the  force  has 
always  been  taken  perpendicular  to  the  axis  of  the  beam  and 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES 


117 


parallel  to  the  section;  but  a  bending  moment  at  a  section  may 
be  caused  by  any  force  that  has  a  lever  arm  with  respect  to  one 
of  the  axes  of  the  section.  Thus  in  Fig.  42  the  force  F  not  only 
has  a  bending  moment  with  respect  to  any  section  such  as  AA 
in  the  horizontal  portion  of  the  beam  but  it  also  has  one  with 
respect  to  any  section  such  as  BB  in  the  vertical  part.  The 
bending  moment  with  respect  to  the  section  BB  is  Fl  where  I  is 
the  perpendicular  distance  in  inches  between  the  action  line  of 
the  force  and  the  axis  of  BB  perpendicular  to  the  plane  of  the  paper.* 
If  a  number  of  forces  parallel  to  F  act  on  the  beam  each  having 
a  lever  arm  with  respect  to  the  same  axis  of  section  BB,  each  will 
cause  a  bending  moment  at  the  section  and  the  algebraic  sum  of 
all  the  bending  moments  will  be  the  total  bending  moment  at 
the  section. 


c 


v 


A 


\,n 


B 


T 


G 


A 


h     j) 


H 


L 


K 


General  Case  of  Bending  Moments.  —  When  a  beam  is  in 
equilibrium  under  the  forces  acting  upon  it  and  all  the  bending 
forces  either  to  the  left  or  to  the  right  of  a  section  are  known,  the 
total  bending  moment  at  that  section  is  obtained  by  taking  the 
algebraic  sum  of  the  individual  bending  moments  at  the  section 
due  to  each  of  the  forces  on  its  left  or  to  each  of  the  forces  on  its 
right.  The  reactions  between  the  beam  and  its  supports  must  be 
included  among  the  forces  considered  and  when  this  can  be  done 
it  is  unnecessary  to  consider  whether  the  beam  is  a  cantilever,  or 


The  force  F  also  produces  tension  in  Section  BB. 


118  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

one  supported  or  fixed  at  both  ends.  If  on  either  side  of  the  sec- 
tion there  are  no  bending  forces  there  is  no  bending  stress  in  the 
section.  Thus  in  Fig.  43  AB  is  a  beam  with  a  projecting  ver- 
tical part,  which  is  in  equilibrium  under  the  action  of  the  forces 
F,  G,  H,  I,  J,  K,  and  L;  F,  G,  and  H  acting  on  the  left  of  the 
section  CD  and  the  others  on  its  right.  The  bending  moment 
at  section  CD,  considering  moments  that  produce  compression 
in  the  upper  fibres  as  positive,  is 

M  =  -Ff  +  Gg-Hh=  -H  +  Jj  -  LI 

The  force  K  does  not  appear  in  the  expression  for  the  bending 
moment  as  it  intersects  the  neutral  axis  of  the  section,  shown 
projected  at  n,  and,  therefore,  has  no  lever  arm  with  respect  to 
it.  In  the  ordinary  case  the  forces  F,  G,  J,  L,  and  K  would  be 
the  reactions  of  the  supports  on  the  beam. 


COMBINED  STRESSES. 

73.  —  It  frequently  happens  in  engineering  structures  that  a 
piece  is  subjected  to  tension  or  compression,  shear,  bending,  and 
torsion,  or  to  some  of  these  stresses,  at  the  same  time.  When 
this  is  the  case  each  stress  is  computed  as  it  if  were  the  only  one 
existing  in  the  piece  and  then  all  are  combined  in  a  manner  to  be 
described. 

Tension  or  Compression  and  Bending.  —  Since  a  bending 
stress  is  one  of  tension  on  one  side  of  the  neutral  axis  of  a  section 
and  compression  on  the  other  side,  a  stress  of  tension  or  com- 
pression in  a  section  due  to  a  simple  tensile  or  compressive  force 
is  added  algebraically  to  the  stress  in  the  section  caused  by  a 
bending  force.  Thus  if  a  piece  subjected  to  a  bending  stress  is 
also  subjected  to  a  simple  stress  of  tension,  the  intensity  of  the 
stress  in  all  the  fibres  on  the  tension  side  of  the  neutral  axis  due 
to  the  bending  force  will  be  increased  by  the  intensity  of  the  simple 
stress  of  tension;  and  the  intensity  of  the  stress  in  all  the  fibres 
on  the  compression  side  of  the  neutral  axis  due  to  the  bending 
force  will  be  decreased  by  the  intensity  of  the  simple  stress  of 
tension. 

Tension  or  Compression  and  Shear.  —  If  a  piece  be  subjected 
to  either  tension  or  compression  and  shear  the  resulting  stress  of 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES  119 

tension  or  compression  in  any  section  due  to  combination  with 
the  shear  is 


Slc  =  |  Sl  +  VS.2  +  GSf,/2)»  (24)* 

and  the  resulting  shearing  stress  due  to  combination  with  the 
tension  or  compression  is 


S.e  =  VS.2  +  (Si/2?  (25) 


in  which  Sic  and  Sfc  are  the  resulting  stresses  per  unit  of  area  of 
tension  or  compression  and  shear,  respectively;  Si  is  the  stress 
per  unit  of  area  of  tension  or  compression  computed  as  if  the 
shearing  stress  did  not  exist;  and  S,  is  the  shearing  stress  per 
unit  of  area  computed  as  if  the  stress  of  tension  or  compression 
did  not  exist.  If  Si  is  tension  Sic  will  be  tension;  and  if  Si  is 
compression  Sic  will  be  compression. 

Bending  and  Shear.  —  Since  a  bending  stress  is  one  causing 
tension  in  the  fibres  on  one  side  of  the  neutral  axis  and  com- 
pression on  the  other  side,  the  formulas  for  combining  it  with  a 
shearing  stress  are  the  same  as  for  combining  a  stress  of  tension 
or  compression  with  one  of  shear.  Si  in  this  case  would  repre- 
sent the  stress  of  tension  or  compression  due  to  the  bending  force 
computed  as  if  the  shearing  stress  did  not  exist. 

Ordinarily  a  bending  stress  is  not  combined  with  a  shearing 
stress  due  to  the  same  bending  force  or  to  a  parallel  bending 
force,  because  such  a  shearing  stress  varies  from  a  maximum  at 
the  neutral  axis,  where  the  bending  stress  is  zero,  to  zero  at  the 
fibre  farthest  from  the  neutral  axis,  where  the  bending  stress  is  a 
maximum. 

Shear  and  Torsion.  —  A  torsional  stress  is  essentially  one  of 
shear  differing  from  the  ordinary  shearing  stress  only  in  that  its 
intensity  per  unit  of  area  varies  with  the  distance  of  the  fibre 
being  considered  from  the  axis  of  the  p  ece  under  torsion.  A 
shearing  stress  distributed  uniformly  over  a  cross-section  may, 
therefore,  be  combined  with  a  torsional  stress  by  simple  algebraic 
addition.  If  the  particular  fibre  under  consideration  is  sub- 
jected to  a  tors  onal  stress  which  acts  in  a  direction  opposite  to 
that  of  the  shear,  the  combined  stress  is  numerically  equal  to 


*  From  Fiebeger's  Civil  Engineering. 


120  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

the  difference  of  the  torsional  and  the  shearing  stresses  computed 
as  if  each  existed  alone,  but  if  at  the  particular  fibre  under  con- 
sideration the  stresses  act  in  the  same  direction  the  combined 
stress  is  numerically  equal  to  their  sum. 

Tension  or  Compression  and  Torsion.  —  Since  a  torsional 
stress  is  a  form  of  shearing  stress  the  formulas  for  combining  it 
with  a  stress  of  tension  or  compression  are  the  same  as  for  com- 
bining a  shearing  stress  therewith.  Sa  in  this  case  would  represent 
the  torsional  stress  computed  as  if  the  stress  of  tension  or  com- 
pression did  not  exist. 

Bending  and  Torsion.  —  The  formulas  for  combining  bend- 
ing and  shear  apply  here  also;  but  since  each  stress  in  this  case 
varies  in  intensity  with  the  position  in  the  section  of  the  fibre 
under  consideration,  care  must  be  taken  to  see  that  the  stresses 
combined  pertain  to  the  same  fibre.  The  maximum  stresses  are 
what  it  is  generally  desired  to  determine  and  consequently  it  is 
customary  to  obtain  first  the  maximum  simple  stress  Si  due  to 
bending  alone,  which  occurs  in  that  fibre  most  distant  from  the 
neutral  axis,  and  to  combine  it  with  the  stress  Sa  due  to  torsion 
alone  which  occurs  in  the  same  fibre.  Ordinarily  the  maximum 
simple  stresses  Si  and  S,  occur  at  the  same  fibre  but  when  this 
is  not  the  case  the  maximum  value  of  St  must  also  be  found  and 
combined  with  the  value  of  Si  at  the  same  fibre  to  determine 
which  of  the  combined  stresses  is  the  greater.  It  may  also 
happen  that  the  maximum  combined  stress  Sic  or  Sac  occurs  in  a 
fibre  in  which  neither  Si  nor  Ss  is  a  maximum.  No  general  rule 
can  be  laid  down  to  cover  a  case  of  this  kind  but  consideration 
of  the  shape  of  the  section  will  generally  indicate  where  the 
maximum  combined  stress  is  likely  to  occur. 

Method  of  Combining  Stresses.  —  If  a  piece  is  subjected  to 
tension  or  compression,  bending,  shear,  and  torsion,  at  the  same 
time,  the  stress  of  tension  or  compression  should  be  combined  by 
algebraic  addition  with  the  bending  stress,  and  the  shearing  stress 
combined  with  the  torsional  stress  in  the  same  way.  The 
resulting  stress  of  tension  or  compression  should  then  be  com- 
bined with  the  resulting  shearing  stress  by  equations  (24) 
and  (25). 


Mfth    • 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES  121 

NEUTRAL  AXIS.      CENTER  OF  GRAVITY.      MOMENT 
OF    INERTIA. 

74.  Neutral  Axis.  —  When  the  bending  force  is  parallel  to 
the  plane  of  the  section  the  neutral  axis  is  the  straight  line  through 
the  center  of  gravity*  of  the  section  perpendicular  to  the  force. 
When  the  bending  force  is  perpendicular  to  the  plane  of.  the  sec- 
tion it  tends  to  bend  it  about  every  straight  line  passing  through 
its  center  of  gravity  that  does  not  intersect  the  action  line  of  the 
force,  prolonged  if  necessary.  Any  such  line,  therefore,  becomes 
the  neutral  axis  of  the  section  when  the  bending  about  itself  is 
being  considered. 

Irregular  Sections.  —  The  positions  of  the  centers  of  gravity 
of  a  number  of  plane  figures  and  the  moments  of  inertia  of  these 
figures  with  respect  to  various  axes  passing  through  their  centers 
of  gravity  are  given  in  most  works  on  mechanics;  but  in  many 
or  most  instances  in  the  design  of  gun  carriage  parts  the  sections 
upon  which  bending  and  torsional  stresses  occur  are  of  complex 
shape  for  which  the  position  of  the  center  of  gravity,  and  the 
moment  of  inertia,  must  be  specially  determined. 

Determination  of  the  Centers  of  Gravity  of  Irregular  Sections. 
-The  position  of  the  center  of  gravity  of  an  irregular  plane 
figure  may  be  determined  from  the  principle  that  the  moment 
of  the  total  area  of  the  figure  about  any  axis  must  equal  the  sum 
of  the  moments  about  that  axis  of  the  various  partial  areas  into 
which  the  figure  may  be  divided.  The  moment  of  an  area  about 
an  axis  is  the  product  of  the  area  by  the  distance  between  the  axis 
and  a  line  parallel  to  it  passing  through  the  center  of  gravity  of 
the  area.  If,  therefore,  the  'sum  of  the  moments  of  the  partial 
areas  about  any  axis  is  determined  the  quotient  obtained  by 
dividing  that  sum  by  the  total  area  will  be  the  distance  from  the 
axis  to  a  parallel  line  passing  through  the  center  of  gravity  of  the 
figure.  By  repeating  these  operations  with  respect  to  another 
axis  the  direction  and  position  of  a  second  line  passing  through 
the  center  of  gravity  of  the  figure  can  be  determined.  The  inter- 


*  What  is  here  called  the  center  of  gravity  of  a  section  is  really  its  center  of 
figure,  for  strictly  speaking  a  surface  has  no  weight.  However,  as  it  is  con- 
venient to  use  the  term  center  of  gravity  for  center  of  figure  this  practice  will  be 
continued  throughout  the  text. 


122 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


section  of  the  two  lines  so  determined  must  necessarily  be  the  center 
of  gravity  of  the  figure. 

When  finding  the  center  of  gravity  of  an  irregular  plane  figure 
or  section  in  this  way,  the  partial  areas  into  which  it  is  sub- 
divided should  be  of  such  regular  forms  that  the  area  and  the 
position  of  the  center  of  gravity  of  each  are  known  or  can  readily 
be  obtained  by  the  ordinary  rules  of  mensuration.  The  areas 
and  the  positions  of  the  centers  of  gravity  of  a  number  of  regular 
plane  figures  are  given  in  Table  5,  in  which  will  also  be  found  the 
moment  of  inertia  of  each  figure  about  an  axis  passing  through 
its  center  of  gravity.  The  last  will  be  needed  for  determining 
the  moments  of  inertia  of  irregular  plane  sections  as  will  be 
described  later. 

Example.  —  To  illustrate  the  method  of  determining  the 
centers  of  gravity  of  irregular  sections,  let  it  be  required  to  de- 
termine the  center  of  gravity  of  the  section  shown  in  Fig.  44. 

fx=1.56+ 


-CENTER  OF  GRAVITY 


1 

i 
i 

•~r 

i 

«-70-* 

i 

s:H 

2.0 
i 
1 

I—  -»~ 

I 

<M 
•<*• 
<M 
c^ 

K\    '!* 

^ 

y ;__| ! I ^._JL 1 


•-I 

k/.0 

Y 


i     •  i 

-»«---2.0---H 


Fig.  44. 


Divide  the  section  into  a  semi-circle,  a  rectangle,  and  a  triangle 
by  the  dotted  lines  as  shown.  Draw  two  axes  XX  and  YY  per- 
pendicular to  each  other  through  the  lowest  and  extreme  left- 
hand  points,  respectively,  of  the  section.  Now  if  two  lines  pass- 
ing through  the  center  of  gravity  of  the  section  be  determined, 
'one  of  them  parallel  to  XX  and  the  other  to  YY,  their  inter- 
section will  give  the  center  of  gravity  desired.  Taking  moments 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES 


123 


TABLE  5. 

In  the  table  the  moment  of  inertia  is  given  for  the  axis  shown  by  the  dotted 
line  passing  through  the  center  of  gravity  of  the  figure.  The  distance  a  is 
measured  from  the  axis  to  some  prominent  line  or  point  of  the  figure. 


Figure. 


Moment  of  inertia.  Distance  a. 


Area  of  figure. 


.2      J' 

£    a 

i_ 


Q, 


bd*/12 


36  (6+60 


-h3 


.110  r4 


.055  r* 


d/2 


h/2 


h/3 


6+2  61  h 
6+61  3 


d/2 


.4244  r 


.4244  r 


6d 


bh 


bh/2 


6+6 


l-h 


=  1.5708  r2 


=  1.5708r2 


124 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


about  XX,  and  consulting  Table  5  for  the  areas  of  the  elementary 
figures  and  the  locations  of  their  centers  of  gravity,  we  have 

TT  x  (l-5)2(>4244  x  1.5  +  2  +  1.5)  +  (1x2)  (1+1.5)  + 

(26) 


in  which  y  is  the  distance  between  the  axis  XX  and  a  line  parallel 
thereto  passing  through  the  center  of  gravity  of  the  section. 
From  equation  (26)        y  =  3.242  ins. 


•y 


X 


Taking  moments  about  YY,  and  adding  the  triangle  shown  in 
double  dotted  lines  in  Fig.  45  to  aid  in  obtaining  the  moment  of 
the  original  triangle  about  this  axis,  we  have 

7T    X    (1.5)2 


x  1.5  +  (1x2)  (.5  +  1) 


(l. 


5  x        l  +  2  - 


(27) 


in  which  x  is  the  distance  between  the  axis  YY  and  a  line  parallel 
thereto  passing  through  the  center  of  gravity  of  the  section. 

From  equation  (27)        x  =  1.56  ins. 

The  coordinates  of  the  center  of  gravity  of  the  figure  with  respect 
to  the  axes  XX  and  77  are,  therefore, 

y  =  3.242  ins.;    x  =  1.56  ins. 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES  125 

Determination  of  the  Moments  of  Inertia  of  Irregular  Sections. 

—  The  moment  of  inertia  of  an  irregular  section  about  any  axis 
in  its  plane  is  the  sum  of  the  moments  of  inertia  of  the  various 
partial  areas  into  which  it  may  be  divided  about  the  same  axis. 
Each  partial  area  should  be  of  such  regular  form  that  its  moment 
of  inertia  about  an  axis  passing  through  its  center  of  gravity 
is  known  or  can  readily  be  obtained  by  the  methods  of  the  cal- 
culus. Having  the  moment  of  inertia  about  an  axis  passing 
through  the  center  of  gravity  of  a  figure,  its  moment  of  inertia 
about  any  other  axis  parallel  to  the  first  is  obtained  by  adding 
the  area  of  the  figure  multiplied  by  the  square  of  the  distance 
between  the  two  axes  to  the  moment  of  inertia  of  the  figure  about 
the  axis  passing  through  its  center  of  gravity. 

The  moments  of  inertia  of  a  number  of  plane  figures  of  regular 
form  about  axes  passing  through  their  centers  of  gravity  are  given 
in  Table  5. 

Example.  —  To  illustrate  the  method  of  determining  the 
moment  of  inertia  of  an  irregular  section  let  it  be  required  to 
determine  the  moment  of  inertia  of  the  section  shown  in  Fig.  44, 
(a)  about  the  axis  through  its  center  of  gravity  parallel  to  XX, 
(6)  about  one  through  its  center  of  gravity  parallel  to  YY,  and 
(c)  about  the  axis  passing  through  its  center  of  gravity  perpen- 
dicular to  its  plane.  The  moment  of  inertia  about  the  last  axis 
is  called  the  polar  moment  of  inertia  and  would  be  used  in  de- 
termining the  torsional  stress  in  the  section. 

Consulting  Table  5  we  find  that  the  moment  of  inertia  of  the 
semicircular  part  of  Fig.  44  about  an  axis  parallel  to  XX  passing 
through  the  center  of  gravity  of  the  section  is 

.110  x  (1.5)4  +  1.5708  x  (1.5)2  x  \  1.5  +2  +.4244x1.5-  3.242  j2 
=  3.385  ins.4 

Similarly,  the  moment  of  inertia  of  the  rectangular  part  about 
the  same  axis  is 

1  X10(2)S  +  {1  x  2  |j  3.242  -  (1.5  +  I)}2  =  1.768  ins.4 
iz 

and  the  moment  of  inertia  of  the  triangular  part  about  the  same 

axis  is 


x         3.242  -    1.5  -  -  3.864  ins.' 

do 


126  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

The  moment  of  inertia  of  the  section  about  this  axis  is,  therefore, 
3.385  +  1.768  +  3.864  =  9.017  ins.4 

The  moments  of  inertia  of  the  partial  areas  of  the  section  about 
an  axis  parallel  to  YY  passing  through  the  center  of  gravity  of 
the  section  are: 
Of  the  semicircle, 

.3937  x  (1.5)4  +  1.5708  x  (1.5)2  x  (1.56  -  1.50)2  =  2.006  ins.4 

Of  the  rectangle, 

9  v  ("H3 
-^^-  +  11  X2J11.56-  (l+.5)}2=.174ins.4 

Of  the  triangle,  (see  Fig.  44.) 

1.5  X  (2)3      |        1.51  |        /        2\  I2     fl.5  x  (1)3 

~36~      M      •~2\\l  hV2-gr  1-56J-L gg- 

+  il  X  ^  }  1  +  (2  -  i)  -  1.56J2]  =  .270  ins.4 
I  r  J          \         w  j  -I 

The  moment  of  inertia  of  the  section  about  this  axis  is,  therefore, 
2.006  +  .174  +  .270  =  2.45  ins.4 

Having  obtained  the  moments  of  inertia  of  the  section  about 
two  axes  in  its  plane  perpendicular  to  each  other  and  passing 
through  its  center  of  gravity,  the  polar  moment  of  inertia  Ip  may 
be  obtained  from  the  principle  that  the  sum  of  the  moments  of 
inertia  of  a  plane  figure  about  two  axes  in  its  plane  perpendicular 
to  each  other  is  the  moment  of  inertia  of  the  figure  about  an  axis 
perpendicular  to  its  plane  passing  through  the  point  of  intersec- 
tion of  the  first  two  axes.  Whence 

Ip  =  9.017  +  2.45  =  11.467  ins.4 

Cubical  Contents,  Centers  of  Gravity,  and  Moments  of  In- 
ertia, of  Irregular  Volumes.  —  The  cubical  contents  and  the 
positions  of  the  centers  of  gravity  of  a  number  of  regular  volumes, 
and  the  moments  of  inertia  of  the  volumes  about  axes  passing 
through  their  centers  of  gravity,  may  be  found  tabulated  in  most 
books  on  mechanics.  To  determine  these  data  for  irregular 
volumes  the  procedure  is  the  same  as  for  irregular  plane  sections. 
The  irregular  volume  is  divided  into  a  number  of  partial  volumes 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES  127 

of  regular  form  or  approaching  some  regular  form  sufficiently 
closely  for  the  purpose.  The  cubical  contents  of  the  whole  is 
then  equal  to  the  sum  of  the  cubical  contents  of  the  various 
partial  volumes;  and  the  position  of  the  center  of  gravity  of  the 
whole  may  be  obtained  by  the  principle  of  moments.  The 
moment  of  inertia  of  the  irregular  volume  about  any  axis  is  equal 
to  the  sum  of  the  moments  of  inertia  about  that  axis  of  the  various 
partial  volumes  into  which  it  is  divided. 

When  the  moment  of  inertia  of  a  volume  about  an  axis  passing 
through  its  center  of  gravity  is  known,  its  moment  of  inertia 
about  any  other  axis  parallel  to  the  first  is  obtained  by  adding 
the  cubical  contents  of  the  volume  multiplied  by  the  square 
of  the  distance  between  the  two  axes  to  the  moment  of  inertia 
of  the  volume  about  the  axis  passing  through  its  center  of 
gravity. 

Weight  and  Mass  of  any  Volume  Composed  of  a  Given  Ma- 
terial. Moment  of  Inertia  of  the  Mass  in  any  Volume.  —  The 
weight  of  a  volume  of  any  material  is  equal  to  the  cubical  contents 
of  the  volume  multiplied  by  the  weight  of  a  unit  volume  of  the 
material  of  which  it  is  composed. 

The  mass  in  a  volume  of  any  material  is  equal  to  the  cubical 
contents  of  the  volume  multiplied  by  the  number  of  units  of  mass 
in  a  unit  volume  of  the  material  of  which  it  is  composed;  or  the 
mass  in  a  volume  of  any  material  is  equal  to  the  weight  of  the 
volume  divided  by  the  acceleration  due  to  the  force  of  gravity. 

The  moment  of  inertia  about  any  axis  of  the  mass  in  a  volume 
of  any  material  is  equal  to  the  moment  of  inertia  of  the  volume 
about  that  axis  multiplied  by  the  number  of  units  of  mass  in  a 
unit  volume  of  the  material  of  which  it  is  composed. 

75.  Permissible  Stresses  in  Gun  Carriage  Parts.  —  Printed 
specifications  are  published  by  the  Ordnance  Department,  U.  S. 
Army,  in  which  are  given  the  elastic  limit  in  tension,  the  tensile 
strength,  the  elongation  per  unit  of  length  at  rupture,  and  the 
contraction  of  area  at  rupture  demanded  by  the  department  for 
the  most  important  materials  used  for  parts  of  gun  carriages. 
In  some  materials  such  as  cast  iron,  copper,  and  bronze,  the 
elastic  limit  is  not  clearly  defined  in  the  testing  machine  and  on 
this  account  it  is  omitted  in  the  specifications.  Sometimes  also 
the  contraction  of  area  at  rupture  is  omitted. 


128 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


The  elastic  limit  in  compression  is  assumed  to  be  equal  to  the 
elastic  limit  in  tension,  and  the  elastic  limit  in  shear  is  taken  as 
four-fifths  of  the  elastic  limit  in  tension. 

When  the  elastic  limit  is  specified  for  a  given  material  the  di- 
mensions of  the  part  to  be  made  thereof  should  be  so  regulated 
that  the  stresses  developed  in  it  shall  not  exceed  one-half  of  the 
elastic  limit.  When  the  tensile  strength  only  is  specified  the  di- 
mensions of  the  part  should  be  such  that  the  stresses  therein 
shall  not  exceed  one-fifth  of  the  tensile  strength.  The  object 
of  prescribing  a  certain  percentage  of  elongation  per  unit  of 
length  and  of  contraction  of  area  at  rupture  for  materials  used  in 
parts  of  gun  carriages  is  to  prevent  their  being  unduly  brittle 
and,  therefore,  likely  to  break  under  shock. 


6.0 -> 


Fig.  46. 

CALCULATION  OF  STRESSES  IN  PARTS  OF   GUN   CARRIAGES. 
3-INCH  FIELD   GUN. 

76.  Bending  and  Shearing  Stresses  in  the  Front  Clip  of  the 
Gun.  —  The  front  clip  of  the  gun  is  shown  in  Fig.  46.  The 
fcrce  A  =  1078  Ibs.,  see  Figs.  15  and  16,  acts  vertically  downward 
on  the  clip.  It  is  the  resultant  of  two  equal  forces  PI  and  P2 
whose  action  lines  are  normal  to  the  surfaces  S  of  the  clip  as 
shown  in  the  figure.  The  parts  of  the  clip  on  which  the  forces 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES  129 

PI  and  Pz  act  make  an  angle  of  22  £°  with  the  horizontal  and  the 
intensities  of  Pi  and  P2  are,  therefore, 


These  parts  of  the  clip  are  .81  in.  long  and  each  may  be  consid- 
ered as  a  cantilever  over  which  the  force  is  uniformly  distributed, 
and  which  tends  to  break  under  the  action  of  the  force  through 
the  section  66  shown  in  the  figure.  The  bending  moment  at  this 
section  is,  from  equation  (19), 

„  ,      583.4  x  (.81)2      ooc  0  . 

M  =  -     01  v/0    '   =  236.3  in.  Ibs. 

.ol  X  Z 

Section  66  is  6  ins.  wide  and  .3  ins.  deep  and  its  neutral  axis 
is  midway  between  the  top  and  the  bottom.  /,  the  moment  of 
inertia  of  the  section  with  respect  to  its  neutral  axis,  is 

6d3/12  =  6  x  (.3)3/12  =  .0135  in.4 
and  y  =  .15  in. 

The  stress  in  the  fibre  most  distant  from  the  neutral  axis  is,  there- 
fore, from  equation  (6), 

S'"y  =  236.3  x  .15/.0135  =  2626  Ibs.  per  sq.  in. 

tension    at    the   top   of    the    section   and   compression   at  the 
bottom. 

Considering  the  shearing  stress  as  uniformly  distributed  over 
section  66  its  intensity  per  unit  of  area  is 

583.4      00.  1U 

^  —  5  =  324  Ibs.  per  sq.  in. 

The  force  Pi  or  P2  also  causes  a  combined  bending  and  tensile 
stress  in  the  part  of  the  clip  above  section  66  where  the  dimension 
.375  is  placed.  The  bending  moment  with  respect  to  the  section 
there  may  be  obtained  by  considering  the  force  as  concentrated 
at  the  middle  of  the  length  of  .81  in.,  whence 


M  =  583.4        +          =  345.7  in.  Ibs. 


130  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

This  section  is  6  ins.  wide  and  .375  in.  deep  and  its  moment  of 
inertia  with  respect  to  its  neutral  axis,  half  way  between  the  inner 
and  outer  edges  of  the  section,  is 

/  =  6  x  (.375)3/12  =  .0264  in.4 
The  maximum  bending  stress  is,  therefore, 

345.7  x  .375/2      O.r_  „ 
S'"y  =  -  -  =  2455  Ibs.  per  sq.  in. 


tension  on  the  inside  and  compression  on  the  outside  of  the  sec- 
tion. 

The  tensile  stress  in  the  section  is 

583.4 


6  x  .375 


=  259  Ibs.  per  sq.  in. 


The  maximum  stress  in  the  section  is,  therefore,  one  of  tension 
which  occurs  on  the  inner  edge  and  is  equal  to 

2455  +  259  =  2714  Ibs.  per  sq.  in. 

So  far  as  the  stresses  in  the  clip  are  concerned  its  length  might 
safely  be  reduced  to  .75  in.  since  the  elastic  limit  of  the  material 
of  which  it  is  made  is  required  to  be  not  less  than  53000  Ibs.  per 
sq.  in.  But  when  one  part  slides  upon  another  as  in  this  case 
it  is  desirable  to  keep  the  pressure  between  the  sliding  surfaces  as 
low  as  possible  to  prevent  wear  and  consequent  loose  fitting  of 
the  parts  after  they  have  been  in  service  for  a  considerable  time. 

STRESSES  IN  THE  RECOIL  LUG  OF  THE  GUN. 

77.  Maximum  Intensity  of  the  Force  P.  Force  Required  to 
Accelerate  the  Recoil  Cylinder.  —  When  the  gun  is  fired  the  recoil 
lug  draws  the  cylinder  filled  with  oil  to  the  rear  compressing  the 
counter-recoil  spring  and  reducing  its  length  from  70  to  25  inches. 
The  resistance  to  drawing  the  cylinder  to  the  rear  including  the 
compression  of  the  spring  is  the  force  P,  see  Figs.  15  and  16, 
which  when  the  gun  is  at  its  extreme  position  to  the  rear  during 
recoil  was  found,  equation  (32),  page  50,  to  be  3849  Ibs.,  corre- 
sponding to  a  total  resistance  to  recoil  of  4077  Ibs. 

Evidently,  however,  the  value  of  P  will  be  considerably  greater 
than  this  when  the  gun  is  just  commencing  to  recoil  for  then  the 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES 


131 


total  resistance  to  recoil,  equation  (12),  page  44,  is  4923  Ibs. 
The  larger  value  of  P  will,  therefore,  be  calculated  and  used  in 
determining  the  stresses  in  the  recoil  lug.  Fig.  47  shows  the 
forces  on  the  gun  when  it  is  at  15°  elevation  and  just  commencing 
to  recoil.* 


Fig.  47. 

Taking  moments  about  the  center  of  mass  of  the  gun,  the  equa- 
tions expressing  the  relations  between  the  forces  (see  article  37) 
are  as  follows: 

P  -  835  sin  15°  +  .15  (A  +  B)  =  Rt  =  4923  (28) 

A  -  B  +  835  cos  15°  =  0  (29) 

Px7.156+.15(A+B)x3.65-5x34.9  -  Ax44.725  =  0    (30) 

From  which 

A  =  101  Ibs.         (31)  F  =  151  Ibs.  (33) 

B  =  908  Ibs.         (32)  P  =  4988  Ibs.  (34) 

It  will  be  seen  that  at  the  beginning  of  recoil  the  force  P  is  con- 
siderably greater  than  at  the  end,  but  that  the  contrary  is  the 
case  with  the  forces  A,  B,  and  F. 

The  force  P,  however,  is  not  the  only  one  acting  on  the  recoil 
lug  of  the  gun,  for  at  the  instant  of  maximum  powder  pressure  the 
recoil  cylinder  being  required  to  move  with  the  gun  to  the  rear 

*  In  determining  the  forces  brought  into  action  between  the  gun  and  the 
carriage  by  a  given  total  resistance  to  recoil,  R,  the  force  of  the  powder  gases 
does  not  have  to  be  considered. 


132  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

has  the  same  acceleration  as  the  gun,  and  the  force  required  to 
produce  this  acceleration  of  the  cylinder  must  be  transmitted  to 
it  through  the  recoil  lug.  The  total  weight  of  the  recoiling  parts 
is  960  Ibs.,  and  since  the  maximum  pressure  in  the  gun  is  33,000 
Ibs.  per  sq.  in.  the  total  pressure  tending  to  move  the  gun  to  the 
rear  is  • 

33000  x  TT  (3) 2/4  =  233264  Ibs.* 

and  the  corresponding  acceleration  is 

233264  x  32.16/960  =  7814  ft.  per  sec.  per  sec. 

The  weight  of  the  cylinder  filled  with  oil  is  61.5  Ibs.  and  the  force 
required  to  give  it  this  acceleration  is 

fil  c 

x  7814  =  14944  Ibs. 


32.16 

The  counter-recoil  spring  is  not  capable  of  transmitting  a  force 
sufficient  to  give  a  material  acceleration  to  any  considerable  part 
of  its  mass,  and,  therefore,  at  the  instant  of  maximum  accelera- 
tion of  the  gun  the  first  and  possibly  the  second  and  third  coils 
at  the  front  end  of  the  spring  will  be  very  much  deflected  while 
the  balance  of  the  spring  will  have  undergone  practically  no  dis- 
placement. On  this  account  the  acceleration  of  the  spring  will 
not  be  considered. 

The  total  force  transmitted  through  the  recoil  lug  because  of 
the  force  P  and  the  acceleration  of  the  recoil  cylinder  is,  then, 

4988  +  14944  =  19932  Ibs. 

The  recoil  lug  of  the  gun  and  the  point  of  application  of  this 
force  are  shown  in  Fig.  48.  The  weakest  section  of  the  lug  to 
resist  the  shearing  action  of  the  force  is  AB  and  the  weakest 
section  to  resist  the  bending  moment  is  CD.  The  lever  arm  of 
the  force  with  respect  to  section  CD  is  3  ins. 


*  Theoretically  the  resultant  force  that  causes  the  recoil  of  the  gun  is  the 
total  powder  pressure  diminished  by  the  total  resistance  to  recoil.  The  latter, 
however,  is  so  small  compared  with  the  maximum  powder  pressure  that  it  is 
not  worth  while  to  consider  it  in  computing  the  maximum  acceleration  of  the 
recoiling  parts  of  this  carriage. 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES 


133 


Shearing  Stress  in  Section  AB.  —  This  section  is  shown  in 
Fig.  48.  The  portion  of  the  recoil  lug  below  it  weighs  about  1.18 
Ibs.  and  to  give  this  weight  the  acceleration  of  the  gun  requires 
a  force  of  286  Ibs.,  which  theoretically  should  be  added  to  the 


TH 

1.375 


SECTION  A-B. 


SECTION  C-D 


Fig.  48. 


force  of  19932  Ibs.  to  determine  the  shearing  stress  in  the  section, 
but  on  account  of  its  insignificance  when  compared  with  the 
latter  force  it  will  be  neglected  in  these  computations. 
The  shearing  stress  in  section  AB  is,  therefore, 


«7  nnrro   n 

3.65  x  1.375  =  3972  Ibs.  per  sq.m. 

Bending  Stress  in  Section  CD.  —  This  section  is  shown  in 
Fig.  48.  The  portion  of  the  recoil  lug  below  it  weighs  about  5.9 
Ibs.  and  to  give  this  weight  the  acceleration  of  the  gun  requires 
a  force  of  1434  Ibs.  The  point  of  application  of  this  force  is  at 
the  center  of  gravity  of  the  portion  of  the  lug  below  the  section 
and  its  lever  arm  with  respect  to  the  section  is  about  1.25  ins. 


134 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


While  its  bending  moment  with  respect  to  section  CD  is  relatively 
unimportant  compared  with  that  of  the  force  of  19932  Ibs.  it 
will  nevertheless  be  considered  in  determining  the  bending  stress 
in  the  section.  The  total  bending  moment  at  the  section  is,  there- 
fore, 

19932  x  3  -f  1434  x  1.25  =  61589  in.  Ibs. 


and  the  maximum  bending  stress  is 
61589  x  1.375X2 


6.6  x  (1.375)3/12 


=  29615  Ibs.  per  sq.  in. 


tension  at  the  rear  edge  of  the  section  and  compression  at  the 
front  edge.  The  elastic  limit  of  the  material  in  the  recoil  lug 
is  not  permitted  to  be  less  than  65000  Ibs.  per  sq.  in. 


STRESSES  IN  THE  RECOIL  CYLINDER. 

78.  Tensile  Stress  in  the  Rear  End  of  the  Recoil  Cylinder.  — 

The  method  of  connecting  the  recoil  cylinder  to  the  recoil  lug  of 
the  gun  is  shown  in  Fig.  49.     The  rear  end  of  the  cylinder  is  closed 


4- RECOIL  LUG. 


K--/./9--M 
Fig.  49. 

by  the  cylinder  end  AA  screwed  into  it  as  shown.  A  cylinder 
end  stud  BB  in  turn  screws  into  the  cylinder  end  and  passing 
through  a  hole  in  the  recoil  lug  is  held  thereto  by  the  nut  C. 
The  thinnest  section  of  the  cylinder  occurs  through  the  bottom 
of  the  threads  into  which  the  cylinder  end  is  screwed.  At  this 
section  the  outer  diameter  is  2.95  ins.  and  the  inner  diameter  2.68 
ins.  The  force  of  19932  Ibs.,  consisting  of  the  resistance  P  and 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES  135 

the  force  required  to  accelerate  the  cylinder,  must  be  transmitted 
through  this  section,  and  the  tensile  stress  therein  is,  therefore, 

19932 


7r[(2.95)2  -  (2.68)2]/4 


=  16694  Ibs.  per  sq.  in. 


Shearing  Stress  in  the  Threads  of  the  Cylinder  End.  —  The 

force  of  19932  Ibs.  also  tends  to  shear  the  threads  connecting  the 
cylinder  end  to  the  cylinder.  The  U.  S.  standard  thread  used 
by  the  Ordnance  Department  is  of  the  shape  shown  in  Fig.  50 


and  the  distance  ab  is  seven  times  the  distance  be.  If  shearing 
occurs  along  any  line  SS  above  the  line  TT  at  the  bottom  of  the 
threads,  the  threads  on  both  parts  of  the  pieces  screwed  together 
will  have  to  be  shorn  through  but,  if  it  occurs  along  the  line  TT, 
only  seven-eighths  as  much  metal  will  be  shorn  through  as  in  the 
former  case.  It  will,  therefore,  be  assumed  that  shearing  if  it 
occurs  would  take  place  along  the  bottom  of  the  threads  and, 
since  the  cylinder  end  has  the  smaller  diameter,  that  its  threads 
would  shear  before  those  of  the  cylinder.  The  diameter  at  the 
bottom  of  the  threads  of  the  cylinder  end  is  2.572  ins.  and  the 
length  of  the  threaded  surface  is  1.19  ins.  Of  this  length  only 
seven-eighths  is  effective  on  account  of  the  flats  be  at  the  bottom 
of  the  threads.  The  total  area  resisting  the  shearing  action  of 
the  force  of  19932  Ibs.  is,  therefore,  equal  to  seven-eighths  of 
the  outer  surface  of  a  cylinder  whose  length  is  1.19  ins.  and  outer 
diameter  2.572  ins.,  and  the  shearing  stress  in  the  threads  is 

19932 
.875  x  TT  x  2.572  x  1.19  =  2369  lbs'  per  sq' in' 

Stress  in  the  Walls  of  the  Recoil  Cylinder  Due  to  the  Interior 
Hydraulic  Pressure.  —  The  pressure  in  Ibs.  per  sq.  in.  of  the  oil 
in  the  recoil  cylinder  must  be  such  that  it  will  exert  a  total  pres- 
sure on  the  front  end  of  the  cylinder,  tending  to  prevent  its 


136  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

movement  to  the  rear,  equal  to  P  minus  the  force  exerted  by  the 
counter-recoil  spring. 

The  intensity  of  P  when  recoil  is  just  beginning,  equation  (34), 
is  4988  Ibs.  and  at  this  time  the  force  exerted  by  the  spring  is 
516  Ibs.,  whence  the  total  pressure  on  the  end  of  the  cylinder 
is  4472  Ibs.  The  effective  area  at  the  front  end  of  the  cylinder  is 
equal  to  the  effective  area  of  the  piston,  which  is  the  area  of  a 
circle  whose  diameter  is  that  of  the  piston,  minus  the  sectional 
area  of  the  piston-rod,  minus  the  area  of  the  slots  in  the  piston 


Fig.  51. 

for  the  throttling  bars.     Fig.  51  shows  a  cross-section  of  the 
piston  and  piston-rod. 

The  shaded  part  is  the  effective  area  of  the  piston,  the  unshaded 
central  part  representing  the  piston-rod.  The  effective  area  is 
equal  to 

(2.556)2- (1.375)4-  ^J(2.556)2- (2.0545)4 ?~^  =3.192 sq.ins. 

and  the  pressure  per  sq.  in.  is 

4472/3.192  =  1401  Ibs.  per  sq.  in. 

This  pressure  occurs  in  front  of  the  piston  and  between  it  and 
the  front  end  of  the  cylinder.  The  thinnest  section  of  this  part 
of  the  cylinder  occurs  in  front  of  the  place  where  the  throttling 
bars  begin  and  has  an  outer  diameter  of  2.95  ins.  and  an  inner 
diameter  of  2.5717  ins. 

Applying  the  formula  for  the  maximum  stress  produced  by  the 

application  of  an  interior  pressure  to  a  simple  cylinder  we  obtain 

2(1.28585)2  +  4(1.475)2 

3  [(1.475)2  -  (1.28585)2] 

tension  at  the  inner  surface. 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES  137 

This  stress  is  not  the  resultant  tension  in  the  section  for  the 
force  P  produces  a  tensile  stress  therein,  in  the  direction  of  the 
axis  of  the  cylinder  and  at  right  angles  to  the  stress  produced  by 
the  interior  pressure,  equal  to 

.[(2.95)'  -'(Sto-m/i  =  304°  lbs" Per  Sq' in" 

and,  as  has  been  shown  in  the  deduction  of  the  formulas  used  in 
calculating  the  stresses  in  a  gun,  a  force  of  tension  or  compression 
produces  a  stress  in  a  direction  perpendicular  to  its  action  line 
equal  to  one-third  of  the  stress  produced  in  that  direction  but  of 
opposite  sign.  The  force  P,  therefore,  produces  a  compression  of 
3040/3  =  1013  lbs.  per  sq.  in. 

in  the  direction  of  the  tension  due  to  the  interior  pressure,  reduc- 
ing that  tension  from  10740  to  9727  lbs.  per  sq.  in. 

This  discussion  relates  to  periods  after  the  acceleration  of  the 
gun  and  recoiling  parts  has  ceased  for,  as  has  been  shown,  at  the 
instant  of  maximum  acceleration  the  tensile  stress  in  the  cylinder 
due  to  the  force  required  to  accelerate  it  is  largely  in  excess  of 
that  due  to  the  force  P  or  to  the  interior  pressure. 

STRESSES  IN  THE  CRADLE. 

79.   Maximum  Value  of  the  Forces  Acting  on  the  Cradle.  — 

It  was  shown  in  determining  the  stresses  in  the  recoil  lug  that 
when  the  gun  is  just  commencing  to  recoil  the  force  P  is  consider- 
ably greater  than  it  is  when  recoil  is  about  to  end,  while  the  con- 
trary is  the  case  with  forces  A  and  B.  Before  computing  the 
stresses  in  some  of  the  parts  of  the  cradle,  therefore,  the  forces 
acting  on  it  when  the  gun  is  just  commencing  to  recoil  will  be 
determined  so  that  the  larger  forces  occurring  under  either  one 
of  the  two  conditions  may  be  used  in  the  computations. 

Fig.  52  shows  the  position  of  the  forces  acting  on  the  cradle 
when  the  gun  is  just  commencing  to  recoil,  and  the  values  of  the 
forces  A,  B,  F,  and  P  already  determined  in  article  77,  page  133. 

Taking  moments  about  the  center  of  mass  of  the  cradle,  the 
equations  representing  the  relations  between  the  forces  are 

D  -  E  +  409  cos  15°  -  101  +  908  =  0  (35) 

C  -  409  sin  15°  -  4988  -  151  =  0  (36) 

D  x  2.277  +  E  x  22.848  -  C  x  3.939  -  101  x  49.277 

-  151  x  3.506  -  908  x  30.348  =  0        (37) 


138 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES 


139 


Whence  C  =  5245  Ibs.  (38) 

D  =  1046  Ibs.  (39) 

E  =  2248  Ibs.  (40) 

Stresses  in  Section  1-1  of  the  Cradle.  —  This  section  is  in- 
dicated in  Figs.  16  and  52.  It  is  shown  in  Fig.  53. 

Moment  of  Inertia  of  Section  1-1.  —  As  this  section  is  sub- 
jected to  a  bending  stress  due  to  the  forces  B,  E,  part  of  F  equal 


;  Ne  u  t  ral  Ax  is. 

~\ 


Area  —  o.oi  s<j  ins 
I  =  29.236  ins.4 


J87 


_D 


Section  1-1. 
Fig.  53. 

to  .155,  and  a  component  of  the  weight  of  that  part  of  the 
cradle  in  rear  of  the  section,  it  will  be  necessary  to  determine  its 
moment  of  inertia  about  a  horizontal  axis  passing  through  its 
center  of  gravity.  The  true  shape  of  the  section  is  given  in  the 


140  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

right  half  of  Fig.  53  and,  while  quite  irregular,  it  may  be  divided 
approximately  into  a  number  of  regular  parts  whose  bases  are 
horizontal  as  shown  in  the  left  half  of  the  figure.  Such  approx- 
imations as  these  are  generally  necessary  to  determine  the  mo- 
ments of  inertia  of  irregular  sections.  It  will  be  observed  that  the 
approximate  figures  include  in  some  cases  small  areas  not  found 
in  the  real  section  and  also  that  some  small  areas  included  in  the 
real  section  are  omitted  in  the  approximate  figures.  When  this 
is  necessary  the  areas  lost  and  gained  should  balance  each  other 
both  in  extent  and  position  so  far  as  possible,  in  order  that  the 
moment  of  inertia  computed  from  the  approximate  regular  figures 
shall  vary  as  little  as  possible  from  that  of  the  real  section. 

To  determine  the  position  of  a  horizontal  line  passing  approx- 
imately through  the  center  of  gravity  of  the  real  section  consider 
only  the  left  half  of  the  figure,  since  the  real  section  is  symmetrical 
with  respect  to  its  vertical  axis,  and  take  moments  with  respect 
to  the  horizontal  line  AB. 

Whence 

1.36  x  4^  J3.42  +4?}  +  1-36  x  4r  I3-42  -  ^ 

4    (  6  J  4    (  6 

+1.74  x  4^2.906  +  ^}  +  .41  X  4^2.94  +  -, 

Z    (  6  J  A    { 

+.41  x  .38J2.56  +  ~\  +  .4  x  4£{2.66  +  ~ 

£t  \  &     \  o 


+.59  X  '-  2.56  -         +  -202  x  2-56  X          - 

«  (  6  J  z 

TT  [5.8125  +  2  x  .187]2       ,9/M  v  5.8125  +  2  x  .187 

~16~  2 

TT  (5.8125)2  x  _4244  x  5.8125  =  (Area  Q£  ^  gection)- 
lo  ^ 

The  total  area  of  the  half  section  is 

40  14  14 

1.36  x  .48  +  1.74  x  ^  +  .41  x  ^p  +  .41  x  .38  +  .4  x  ^ 

+.59  x  f  +  .202  x  2.56  +  ^5.8125  +  2  x  .187  P 
=2.806  sq.  ins. 


and  y  =  1.190  ins. 

The  area  of  the  whole  section  is  5.61  sq.  ins. 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES  141 

The  moment  of  inertia  of  the  half  section  with  respect  to  the 
horizontal  axis  passing  through  its  center  of  gravity  is 

1  3fi  v  (AR\3  AQ  f  AQ  12 

%V    j  +  1.36  x  ~  3.42  +^  -  1.190 

OO  6     \  O 


*  I**)3  +  1.74  x        2.906  +        -  1.190'! 


.  . 

OO  Z  O 

41X04)3 

OO 


+  -41  *(•»)•  +  .41  x  .38     56  +        -  1.190'' 


+  A  X9i'14)3  +  -4  X  ~  J2.56  +  ~  -  1.19012 

OO  —  •> 

I 

CQ    v    /  K7U  KS7  (  K7  12 

.o,  xi ^o/£  +  59  x  .o<_  2  56  _  ^.  _  L190 

00  Z     [  O 

909  v  (9  ^fi^3  f9  Kfi 

.     vfAitA   A    ^.OO^       .      OAO    vx  o  tr<?  M-".«O         ^    -\r\f\ 

+  —    — r«j—  -^  +  .202  X  2.56]—^ 1.190 

!  -  Z 


x  [4244  X  5.8125  +2  x  .187  + 

I  Z 

-  ^  (5.8125)2  x  (.4244  x  ^^  +  1.19of=  14.618  ins.4 
lb  I  Z  J 

The  moment  of  inertia  of  the  whole  section  with  respect  to  the 
horizontal  axis  through  its  center  of  gravity  is,  therefore, 

/  =  29.236  ins.4 

Bending  Moment  and  Stress  in  Section  1-1.  —  The  bending 
moment  at  this  section  is  obtained  by  taking  the  algebraic  sum  of 
the  moments  of  the  forces  acting  on  the  right  (to  the  rear)  of  it. 
In  doing  this  it  will  be  found  that  while  the  forces  are  greater 
when  recoil  of  the  gun  is  about  to  end,  the  resultant  bending 
moment  is  slightly  less  than  when  the  gun  is  just  commencing 
to  recoil.  The  stresses  under  the  latter  assumption  will,  there- 
fore, be  computed.  In  the  computations  which  follow  the  weight 


142  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

of  the  cradle  will  be  neglected  as  only  that  part  of  it  in  rear  of 
the  section  produces  a  bending  moment.  This  bending  moment 
is  slight  and  neglecting  it  increases  the  resultant  bending  moment 
at  the  section,  which  is  an  error  on  the  safe  side. 

In  determining  the  bending  moment  it  should  be  noted  that 
only  that  part  of  F  which  is  due  to  the  friction  produced  by  B 
must  be  considered  since  the  part  due  to  the  friction  produced  by 
A  acts  on  the  left  (in  front)  of  the  section.  Furthermore  the 
lever  arm  of  .15  B  must  be  taken  with  reference  to  the  horizontal 
axis  through  the  center  of  gravity  of  the  section  which,  from  Fig. 
53,  is  1.190  ins.  above  the  action  line  of  the  force  P,  or  the  axis  of 
the  lower,  cylindrical,  part  of  the  cradle.  The  bending  moment 
at  the  section  is,  then,  see  Figs.  52  and  53, 

M=#xl8.625-.Bx26.125-.15B  (3.506-1.190)  =17832  in.  Ibs. 

since 

E  =  2248  Ibs.  and  B  =  908  Ibs. 

The  lowest  point  of  the  section  is  at  the  greatest  distance  from 
the  neutral  axis,  this  distance  being 


1.190  +  +  .187  =  4.283  ins. 

Z 

and  the  bending  stress  at  that  point  is 
S'"y  =  17832  x  4.283/29.236  =  2612  Ibs.  per  sq.  in.  tension. 

The  force  .15  B  produces  a  tensile  stress  distributed  uniformly 
over  the  section  equal  to  136/5.61  =  24  Ibs.  per  sq.  in.  which 
increases  very  slightly  the  tension  below  the  neutral  axis  and 
decreases  by  the  same  amount  the  compression  above  it. 

The  forces  B  and  E  produce  a  shearing  stress  in  the  section 
which  is  greatest  when  recoil  is  about  to  end.  At  this  tune  its 
intensity  is,  see  equations  (40)  and  (31),  chapter  II, 

[3511  -  1885J/5.61  =  290  Ibs.  per  sq.  in. 

Stresses  in  Section  2-2  of  the  Cradle.  —  This  section  is 
indicated  in  Figs.  16  and  52.  It  is  practically  identical  in  shape 
and  area  with  section  1-1,  Fig.  53,  and,  therefore,  the  area, 
position  of  neutral  axis,  and  moment  of  inertia  determined  for 
that  section  will  also  serve  for  it.  The  stresses  in  the  section  are 
greatest  when  the  gun  is  just  commencing  to  recoil. 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES  143 

The  forces  P  and  .15  A,  see  Fig.  52,  produce  a  stress  of  com- 
pression uniformly  distributed  over  the  section  which  is  equal  to 

[4988  +  15J/5.61  =  892  Ibs.  per  sq.  in. 

The  section  is  also  subjected  to  a  bending  moment  which  may 
be  found  by  taking  the  algebraic  sum  of  the  moments  of  the  forces 
acting  on  the  left  (in  front)  of  it.  Neglecting  the  weight  of  the 
cradle  as  in  the  case  of  section  1-1  and  noting  by  reference  to  Fig. 
53  that  the  force  P  has  a  lever  arm  of  1.190  ins.  with  respect  to 
the  neutral  axis  of  the  section,  we  obtain,  see  Fig.  52, 

M=Ax43.50+.15A(3.506-1.190)-Pxl.l90=-1507in.  Ibs. 

since 

A  =  101  Ibs.  and  P  =  4988  Ibs. 

The  maximum  bending  stress  in  the  section  occurs  at  its  lowest 
point  and  is 

S'"y  =  1507x4.283X29.236  =221  Ibs.  per  sq.  in.  compression. 

The  latter  stress  increases  the  compressive  stress  uniformly 
distributed  over  the  section  so  that  the  maximum  resultant 
stress  is 

892  +221  =  1113  Ibs.  per  sq.  in.  compression 

and  occurs  at  the  lowest  point  of  the  section. 

The  force  A  also  produces  a  shearing  stress  in  the  section 
which  is  greatest  when  the  gun  is  about  .5  in.  from  the  end  of 
its  recoil.  At  this  time  the  intensity  of  A  is  practically  1078 
Ibs.  and  the  corresponding  shearing  stress  is 

1078X5.61  =  192  Ibs.  per  sq.  in. 

It  will  be  observed  that  the  cradle,  so  far  as  the  stresses  in  it 
due  to  the  firing  of  the  gun  are  concerned,  could  be  made  much 
lighter,  but  its  walls  are  now  only  .187  in.  thick  and  if  made  much 
thinner  they  would  be  likely  to  be  deformed  and  injured  in  rapid 
transportation  or  maneuvering  of  the  battery. 

Stresses  in  the  Pintle  and  in  the  Rivets  fastening  it  to  the 
Cradle.  —  Fig.  54  shows  the  pintle  of  the  3-inch  field  carriage 
with  the  forces  C  and  D  acting  thereon. 

The  pintle  is  fastened  to  the  cradle  by  twenty-three  .375- 
m.  rivets  through  the  rivet  holes  shown  in  the  figure.  The 


144 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES  145 

greatest  value  of  the  force  C  (5245  Ibs.)  occurs  when  the  gun  is 
just  commencing  to  recoil  while  that  of  the  force  D  (2312  Ibs.) 
occurs  when  recoil  is  about  to  end.  These  values  will  be  used 
in  the  computations  that  follow. 

Stresses  Caused  by  the  Force  C.  —  The  lower  projecting  part 
of  the  pintle  against  which  the  force  C  acts  is  practically  cylindri- 
cal in  shape  (it  has  a  very  slight  taper)  and  forms  a  cantilever 
1.56  ins.  long  over  which  the  force  is  uniformly  distributed.  The 
dangerous  section,  see  Fig.  54,  occurs  at  a  distance  of  1.43  ins. 
above  the  lower  edge  of  the  pintle  and  the  bending  moment  there 
is,  from  equation  (18), 

,,      5245  v  (1.43)2      QA9Q .     .. 
M  =  =-=5-  x      0      =  3438  in.  Ibs. 
l.oo          z 

The  dangerous  section  is  a  circular  ring,  its  outer  diameter  being 
approximately  4.998  ins.  and  its  inner  diameter  4.375  ins.  Its 
moment  of  inertia  about  its  neutral  axis  is,  therefore, 

.0491  {(4.998)*  -  (4.3T5)4}  =  12.651  ins.4 
and  the  bending  stress  at  the  extreme  fibre  is 

3438  x  4.998X2      ™  « 
y  =  ~      10651        "  =  per  sq* m< 

compression  at  the  front  of  the  section  and  tension  at  the  rear. 
The  shearing  stress  in  the  section  is 

5245  x  1.43/1.56 


.7854  { (4.99S)2  -  (4.375)2j 


=  1049  Ibs.  per  sq.  in. 


The  force  C  tends  to  shear  the  twenty-three  rivets  fastening 
the  pintle  to  the  main  body  of  the  cradle.  These  rivets  are  .375 
in.  in  diameter  and  consequently  .11045  sq.  in.  in  area.  The 
shearing  stress  in  them  is 

nr\i-*r-    11  * 

=  2065  Ibs.  per  sq.  in. 


23  X.  11045 

C  also  tends  to  crush  the  rivets  and  the  walls  of  the  rivet  holes 
in  the  pintle  and  the  cradle.  As  the  thickness  of  the  cradle  is 
less  than  that  of  the  pintle  the  greater  crushing  stress  will  occur 
between  the  rivets  and  the  walls  of  the  rivet  holes  in  the  cradle. 
The  crushing  stress  between  a  rivet  and  the  walls  of  the  rivet  hole 


146  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

is  considered  to  be  uniformly  distributed  over  an  area  equal  to 
the  thickness  of  the  plate  multiplied  by  the  diameter  of  the  rivet. 
Under  this  assumption  the  crushing  stress  due  to  the  force  C  is 

23  x   187  x  375  =  3252  lbs'  per  sq'  in*  comPression» 

since  there  are  twenty-three  rivets  having  a  diameter  of  .375  in., 
and  the  thickness  of  the  cradle  wall  is  .187  in. 

The  force  C  also  tends  to  rotate  the  pintle  about  the  line  cut 
from  the  surface  of  contact  with  the  cradle  by  a  plane  perpen- 
dicular to  the  axis  of  the  cradle  passing  through  the  center  of  the 
pintle.  This  tendency  to  rotation  is  mainly  resisted  by  the 
compression  between  the  contact  surfaces  of  the  cradle  and  pintle 
in  front  of  the  central  plane,  and  by  the  stress  in  the  rivets  behind 
that  plane.  The  manner  of  distribution  of  the  stress  in  these 
rivets  is  not  unlike  that  of  the  tensile  stress  on  one  side  of  the 
neutral  axis  of  a  section  of  a  beam  subjected  to  a  bending  force. 
The  greatest  stress  will  occur  in  the  rivets  at  the  rear  of  the  pintle 
and  will  be  mainly  a  tensile  stress.  It  can  be  obtained  approxi- 
mately with  the  error  on  the  safe  side  in  the  following  manner: 
Assume  that  the  tendency  of  the  pintle  to  rotate  under  the  action 
of  the  force  C  is  resisted  entirely  by  the  pressure  along  the  line  of 
contact  at  the  front  edge  of  the  pintle  and  by  the  tension  in  the 
five  rear  rivets  marked  T  in  Fig.  54.  Assume  also  that  the  mean 
distance  of  the  five  rivets  from  the  front  edge  of  the  pintle  is  9  ins. 
From  Fig.  54  the  lever  arm  of  C  with  respect  to  the  line  of  contact 
at  this  edge  is  .846  in.  Now  since  the  pintle  is  in  equilibrium, 
the  moments  of  the  forces  about  the  line  of  contact  at  the  front 
edge  must  be  zero  and,  calling  T  the  force  which  the  five  rivets 
oppose  to  the  rotation  of  the  pintle,  we  have 

T  X  9  =  5245  x  .846 
or  T  =  493  lbs. 

and  the  tensile  stress  in  each  of  the  five  rivets  is 


Stresses  Caused  by  the  Force  D/2.  —  The  force  D/2  =  1156 
lbs.  acts  downward  on  the  lower  lip  of  each  wing  of  the  pintle 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES  147 

as  shown  in  Fig.  54.  These  lips  may  be  considered  as  cantilevers 
.562  in.  long,  approximately  3  ins.  wide,  and  .372  in.  deep  with 
the  forces  uniformly  distributed  over  them.  The  area  of  the 
dangerous  section  GG  shown  in  Fig.  54  is  1.116  sq.  ins.  and  its 
moment  of  inertia  with  respect  to  its  neutral  axis  is  .01287  in.4 
The  bending  moment  at  this  section  is,  from  equation  (19), 

1156      (.562)2      00. 

-n  X          '   =  324.8  in.  Ibs. 


and  the  maximum  bending  stress  is 

S'"y  =  324.8  x  .186/.01287  =  4695  Ibs.  per  sq.  in. 

The  shearing  stress  in  the  section  is 

1156/1.116  =  1036  Ibs.  per  sq.  in. 

The  bracket  connecting  the  wing  to  the  rest  of  the  pintle,  see 
Fig.  54,  acts  as  a  triangular  frame  supporting  the  force  D/f2. 
The  upper  arm  of  the  bracket  makes  an  angle  of  45°  with  the 
lower  arm,  and  the  force  is  perpendicular  to  the  lower  arm.  The 
total  stress  in  the  upper  arm  is,  therefore,  one  of  tension  equal  to 

i  D/cos  45°  =  1635  Ibs. 

The  greatest  stress  per  unit  of  area  in  the  upper  arm  occurs  at 
the  smallest  section  thereof  which  is  .25  in.  thick  and  3  in.  wide. 
This  stress  is 

-n=  -  Q  =  2180  Ibs.  per  sq.  in. 
.Zo  X  o 

The  total  stress  in  the  lower  arm  is  one  of  compression  equal  to 
D/2  =  1156  Ibs.  The  area  of  cross-section  of  the  lower  arm 
being 

.25  x  3  =  .75  sq.  in. 

the  stress  per  unit  of  area  is 

1156/.75  =  1541  Ibs.  per  sq.  in. 

The  total  tensile  stress  of  1635  Ibs.  in  the  upper  arm  of  the 
bracket  puts  a  tensile  stress  in  a  number  of  the  rivets  fastening 
the  pintle  to  the  main  body  of  the  cradle.  This  stress  is  mainly 
confined  to  the  four  rivets  near  the  junction  of  the  upper  arm  with 


148 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


the  body  of  the  pintle  and,  assuming  it  to  be  entirely  so  confined 
and  uniformly  distributed  over  the  four  rivets,  its  intensity  is 

4  x    11045  =  3701  Ibs.  per  sq.  in. 

Maximum  Resultant  Stress  in  the  Rivets.  —  To  obtain  the 
maximum  resultant  stress  in  the  rivets  the  simple  stresses  occur- 
ring in  them  should  be  combined  in  accordance  with  the  methods 
already  explained,  but  on  account  of  the  low  intensities  of  the 
simple  stresses  this  is  unnecessary. 

STRESSES   IN   SECTION  3-3  OF  THE  TRAIL. 

80.  Forces  Causing  Stress  in  the  Section.  —  Area  and  Moment 
of  Inertia  of  the  Section.  —  This  section  is  indicated  in  Fig.  18 
and  shown  in  Fig.  55.  It  is  taken  through  the  elevating  screw 

Y 


~f~ 

./875-k, 

-* 

^T"  "t1  '  ^"x  —  0                                   .         <n 

\*»                2 

to                                                                          °0 

^                                          >* 
i, 

'^>                                 7f)K 

t. 
<-                          705 

[    Y          V                                          ^ 

f 

A-4' 
* 

"if 
io 

«M 

^                            Area  of  Half  Section  =  2.OS9  sq.  ins. 
1     ijs^                           Ix  of  Half  Section  =  14.431  ins.4 
/„  of  Half  Section  =  .643  in.4 
.;A,.^.*.^,v^/»              IP  °f  Half  Section  =  15.074  ins.*                                 mwu* 

Section  3-3. 
Fig.  55. 

bearings  and,  as  the  force  /  =  4387  Ibs.  is  contained  in  its  plane, 
it  is  the  dangerous  section  of  the  trail.  The  stresses  in  it  are 
greatest  when  recoil  is  about  to  end  and  they  will  be  determined 
under  that  condition. 

The  stresses  in  the  section,  see  Figs.  18  and  55,  are  compression 
and  bending  due  to  the  force  S  =  3938  Ibs.,  bending  and  shear 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES  149 

due  to  the  force  T  =  2139  Ibs.,  and  torsion  due  to  the  force  I  = 
4387  Ibs.  The  force  /  lies  in  the  plane  of  the  section  and  has  no 
bending  moment  with  respect  to  it.  Because  of  the  torsion  pro- 
duced by  /,  which  is  somewhat  special  and  will  be  explained  later, 
it  will  simplify  the  problem  to  consider  each  half  of  the  symmetri- 
cal section  3-3  by  itself,  in  which  case  the  forces  acting  at  the 
spade  should  be  taken  as 

S/2  =  1969  Ibs.  and  T/2  =  1069.5  Ibs. 

The  area  of  the  half  section  is  2.039  sq.  ins.  Its  center  of 
gravity  is  given  by  the  intersection  of  the  horizontal  and  vertical 
axes  XX  and  YY,  the  former  being  at  a  distance  of  3.625  ins. 
from  the  lower  edge  of  the  section  and  the  latter  at  a  distance  of 
.427  in.  from  the  outer  edge.  The  moment  of  inertia  Ix  about 
the  axis  XX  is  14.431  ins.4,  Iy  =  .643  in.4,  and  Ip  =  15.074  ins.4. 

Stresses  in  Section  3-3  due  to  the  Forces  S  and  T.  —  The 
compression  in  the  half  section  due  to  the  force  S/2  is 

1969/2.039  =  966  Ibs.  per  sq.  in. 

The  lever  arm  of  T/2  with  respect  to  the  half  section,  see 
Fig.  18,  is  . 

112.19  -  21.55  =  90.64  ins. 

and  that  of  S/2  with  respect  to  the  neutral  axis  XX,  see  Figs.  18 

and  55,  is 

4  +  16  +  3.625  =  23.625  ins. 

The  resulting  bending  moment  at  the  half  section  is,  then, 
1069.5  x  90.64  -  1969  x  23.625  =  50422  in.  Ibs. 
and  the  maximum  bending  stress  is 

S"'y  =  50422  x  3.625/14.431  =  12666  Ibs.  per  sq.  in. 

compression  at  the  upper  and  tension  at  the  lower  edge  of  the 
section.  The  shearing  stress  in  the  half  section  due  to  the  force 
T/2  is 

1069.5/2.039  =  525  Ibs.  per  sq.  in. 

Torsional  Stress  in  Section  3-3  caused  by  the  Force  /. — 

Referring  to  Fig.  55,  it  will  be  seen  that  the  force  7,  which  is 
the  thrust  on  the  elevating  screw,  acts  midway  between  the 


150  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

two  flasks  of  the  trail.  It  is  transmitted  to  each  flask  by  two 
cross  transoms  riveted  thereto  which  together  form  a  fixed  beam 
with  the  force  applied  at  its  middle.  The  length  of  the  beam  is 
the  distance  between  the  webs  of  the  flasks  equal  to  14.1  ins. 
The  bending  moment  at  the  ends  of  the  transoms  under  these 
conditions  is  given  by  equation  (11)  and  is 

M  =  4387  x  14.1/8  =  7732  in.  Ibs. 

This  bending  moment  will  produce  a  twisting  couple  of  equal 
value  on  each  flask  which  will  cause  a  torsional  stress  in  the  half 
section  3-3  and  in  the  sections  on  each  side  of  it,  between  it  and 
the  axle  and  between  it  and  the  tool  box.  The  axle  and  the  plates 
forming  the  top,  bottom,  and  front  end  of  the  tool  box  prevent 
twisting  in  the  parts  of  the  flasks  connected  by  them  so  that  the 
right  sections  of  the  flasks  tangent  to  the  axle  at  its  rear  and  those 
at  the  front  end  of  the  tool  box  may  be  considered  as  fixed  ends 
so  far  as  the  torsional  stress  in  the  flasks  is  concerned. 

The  distance  between  the  half  section  3-3  and  the  axle  is 
20.902  ins.  and  between  it  and  the  front  end  of  the  tool  box  the 
distance  is  15.224  ins.,  so  that  the  torsional  moment  in  the  half 
section  is,  from  equation  (2),  ^ 

Mt  =  7732  x  20.902/36.126  =  4474  in.  Ibs. 

and  the  maximum  torsional  stress,  which  occurs  at  the  points  0, 
Fig.  55,  is 


V(3.625)2+  (1.573)2/15.074  =  1173  Ibs.  per 
sq.  in.  shear. 

Combination  of  Compressive,  Bending,  and  Torsional  Stresses 
in  Section  3-3.  —  The  compressive  stress  of  966  Ibs.  per  sq.  in. 
is  uniformly  distributed  over  the  section;  it  therefore  increases 
the  compression  of  12666  Ibs.  per  sq.  in.  at  the  upper  edge  of  the 
section,  due  to  the  bending  stress,  to  13632  Ibs.  per  sq.  in.,  and 
reduces  the  tension  at  the  lower  edge  to  11700  Ibs.  per  sq.  in. 
The  compression  of  13632  Ibs.  per  sq.  in.  being  the  greater  stress, 
it  will  be  combined  with  the  torsional  stress  of  1173  Ibs.  per  sq.  in. 
at  0  to  obtain  the  maximum  combined  stress  in  the  section.  From 
equation  (24)  this  maximum  stress  is  found  to  be 


Slc  =  ™p  +  y/(1173)2  +  0?p2  =  13732  Ibs.  per  sq.  in. 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES 


151 


compression  at  0  on  the  upper  edge  of  the  section.  The  torsional 
stress  at  0  is  so  slight  that  it  will  not  be  necessary  to  determine 
how  much  it  is  increased  by  the  compression  at  that  point. 

STRESSES  IN  PARTS  OF  THE  5-INCH  BARBETTE 
CARRIAGE,  MODEL  OF  1903. 

81.  Stresses  in  the  Trunnions  of  the  Cradle.  —  The  cradle 
may  be  considered  as  a  beam  carried  in  the  trunnion  beds  of 
the  pivot  yoke.  The  diameter  of  the  trunnions  of  the  cradle, 
see  Fig.  56,  is  .01  in.  smaller  than  the  diameter  of  the  trunnion 
beds  and  this  fact,  in  connection  with  the  immense  depth  of  the 
sections  of  the  cradle  in  a  direction  parallel  to  the  action  lines 
of  the  principal  forces  acting  on  it,  makes  of  the  cradle  a  beam 
merely  supported  at  its  ends  instead  of  a  fixed  beam.  A  fixed 
beam  is  one  whose  ends  are  held  so  firmly  by  the  supports  that 
they  cannot  accommodate  themselves  to  any  deflection  of  the 
part  of  the  beam  between  the  supports  caused  by  the  bending 


--  4.99— 


NEUTRAL 
AXIS". 


_        .  T0_ 
CENTER  OF  CRADLE. 


Area  of  Section  AB 


17.15  sq.  ins.  /  of  Section  AB 

Fig.  56. 


:  29.98  ins.* 


forces  acting  upon  it.  A  free  beam  is  one  whose  ends  are  merely 
supported  in  such  manner  that  they  can  freely  accommodate 
themselves  to  the  deflection  of  the  part  of  the  beam  between  the 
supports.  By  the  methods  discussed  in  works  on  strength  of 
materials  it  can  be  shown  that  the  maximum  deflection  of  the 
cradle,  which  occurs  midway  between  the  trunnion  beds,  is  less 
than  .00002  in.  so  that  it  is  entirely  negligible,  and  the  clearance 
between  the  trunnions  and  their  beds  is  many  times  greater  than 
it  need  be  to  permit  the  trunnions  to  conform  to  the  deflection  of 
the  rest  of  the  cradle. 
One  of  the  trunnions  of  the  cradle  is  shown  in  Fig.  56. 


152  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

The  forces  acting  on  each  trunnion,  see  Fig.  21,  are  the  hori- 
zontal force  C/2  =  46647  Ibs.  and  the  vertical  force  D/2  = 
5359  Ibs.  The  resultant  of  these  forces  is 

V(46647)2  +  (5359)2  =  46954  Ibs. 

It  is  uniformly  distributed  over  the  surface  of  contact  between 
the  trunnion  and  its  bed  in  the  pivot  yoke.  This  surface  is  4  ins. 
long,  being  .03  in.  shorter  than  the  trunnion  to  allow  a  clearance 
of  .015  in.  between  the  rimbase  and  the  side  of  the  trunnion  bed 
on  the  inside,  and  between  the  half  collar  on  the  end  of  the  trun- 
nion and  the  side  of  the  trunnion  bed  on  the  outside. 

The  bending  moment  at  section  AB  is  greater  than  at  any 
other  section  of  the  trunnion.  It  may  be  obtained  by  consid- 
ering the  resultant  force  concentrated  at  the  middle  of  the  trun- 
nion length,  whence 

M  =  46954  x  2.015  =  94612  in.  Ibs. 
and  the  corresponding  maximum  bending  stress  is 

94612  x  2.495      _Q_.  „ 
*  = 2998      "  =  per  sq>  m' 

tension  at  the  rear  and  compression  at  the  front  of  the  section. 
The  shearing  stress  in  section  AB  is 

46954X17.15  =  2738  Ibs.  per  sq.  in. 

Stress  in  the  Recoil  Cylinder  of  the  Cradle  due  to  the  Interior 
Hydraulic  Pressure.  —  The  cylinder  liner  in  which  the  recoil 
grooves  are  cut  is  made  in  halves  and  adds  nothing  to  the  strength 
of  the  cylinder,  which  outside  of  the  liner  has  an  inner  diameter 
of  7.5  ins.  and  an  outer  diameter  of  11  ins.  The  force  P  = 
84665  Ibs.,  see  Fig.  21,  is  the  sum  of  the  forces  exerted  by  the 
piston-rod  of  the  recoil  cylinder  and  the  spring  rods  of  the  spring 
cylinders.  The  force  exerted  by  the  spring  rods,  which  is  due  to 
the  springs,  is  least  when  the  gun  is  just  commencing  to  recoil, 
at  which  time  it  has  a  value  of  6540  Ibs.,  so  that  the  correspond- 
ing force  exerted  by  the  piston-rod  of  the  recoil  cylinder  is  78145 
Ibs.  The  piston  of  the  recoil  cylinder  has  a  diameter  of  6.49  ins. 
and  the  rod  a  diameter  of  3  ins.  and  consequently  the  area  against 
which  the  oil  in  the  cylinder  must  act  to  produce  a  total  force  of 
78145  Ibs.  is 

.7854  K6.49)2  -  (3)2|  =  26.01  sq.  ins. 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES 


153 


and  the  corresponding  interior  pressure  per  sq.  in.  in  the  cylinder 
is 

78145/26.01  =  3004  Ibs.  per  sq.  in. 

The  maximum  stress  in  the  cylinder  due  to  this  interior  pressure 
is  one  of  tension  and  occurs  at  its  inner  surface.     It  is  as  follows: 


2(3.75)2  +  4(5.5)2 
3f(5.5)2  -  (3.75)2j 


X  3004  =  9226  Ibs.  per  sq.  in. 


Stresses  in  Section  4-4  of  the  Cradle.  —  This  section,  taken 
just  in  front  of  the  spring  cylinders,  is  indicated  in  Fig.  21  and 
shown  in  Fig.  57. 


Trace  of  Horizontal  Plane  containing  Axis 


Area  =  51.24  sq.  ins.  I  =  729.85  ins.4 

Section  4-4. 
Fig.  57. 

There  is  a  bending  moment  at  the  section,  see  Fig.  21,  which 
may  be  obtained  by  considering  all  the  forces  acting  to  the  right 
(in  rear)  of  it.  The  weight  of  the  cradle  will  be  neglected  as 
only  a  part  of  it  acts  to  the  right  of  the  section  and  neglecting  the 
weight  increases  the  bending  moment  slightly.  The  bending 
moments  of  the  forces  perpendicular  to  the  section  must  be  taken 
with  respect  to  its  neutral  axis  which  is  7.178  ins.  below  the  hori- 
zontal plane  containing  the  axis  of  the  trunnions. 


154  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

Whence 

M  =  84665  (15.75  -  7.178)  +  4662  (10  -  7.178) 

-  31084  (44.875  -  20)  +  4311  cos  14°  (35  cos  14°  -  20) 
+  4311  sin  14°  (35  sin  14°  -  7.178)  =  25433  in.  Ibs. 

and  the  corresponding  maximum  bending  stresses  are 

S'"y  =  25433  x  8.178/729.85  =  285  Ibs.  per  sq.  in. 

compression  at  A  and  A,  Fig.  57,  and 

S'"y  =  25433  x  6.822/729.85  =  238  Ibs.  per  sq.  in. 

tension  at  B,  Fig.  57. 

It  will  be  noted  that  while  the  bending  moments  of  the  indi- 
vidual forces  at  this  section  are  large,  they  nearly  neutralize  each 
other  so  that  the  resultant  bending  moment  is  small  and,  in  con- 
nection with  the  large  moment  of  inertia  of  the  section,  produces 
very  little  bending  stress. 

The  forces  P,  Ff,  and  E  sin  14°  produce  a  tensile  stress  in  the 
section  equal  to 

[84665  +  4662  +  4311  sin  14°]/51.24  =  1764  Ibs.  per  sq.  in. 

The  forces  B  and  E  cos  14°  produce  a  shearing  stress  in  the 
section  equal  to 

[31084  -  4311  cos  14°]/51.24  =  525  Ibs.  per  sq.  in. 

As  the  force  E  is  applied  at  a  distance  of  16.125  ins.  to  the  left 
of  the  center  of  the  section,  its  horizontal  component  causes  a 
bending  stress  in  the  section  about  a  vertical  neutral  axis,  and  its 
vertical  component  causes  a  torsional  stress.  Owing  to  the  large 
moments  of  inertia  of  the  section  these  stresses  will  be  small  and 
will,  therefore,  not  be  computed. 

The  maximum  combined  stress  in  the  section  occurs  at  B  and 
is  2002  Ibs.  per  sq.  in.  tension  obtained  by  adding  the  tension  due 
to  the  bending  stress  and  that  due  to  the  direct  action  of  the 
forces  P,  F',  and  E  sin  14°.  If  the  torsion  be  considered  this 
maximum  stress  will  be  increased  slightly. 

STRESSES  IN   THE  PIVOT  YOKE. 

82.  Stresses  in  Section  5-5  of  the  Pivot  Yoke.  —  This  section 
is  indicated  in  Figs.  22  and  23  and  shown  in  Fig.  58. 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES 


155 


It  is  selected  rather  than  one  at  which  the  bending  moment 
is  greater  because  the  increase  in  the  strength  of  the  sections  as 
we  pass  further  down  is  greater  than  the  increase  of  the  bending 
moment. 

k-MIDDLE  OF  TRUNNION    LENGTH. 

r 

K- 7.25 — M- 


X- 


r*  t 

^ 

1 

.. 

M 

1 

\    i 

He—  4.75—  * 

c. 

-     - 

r£ 

> 

S 

««i 

.5/5- 

* 

" 

T 

j^- 

<—12.5  to  center  of  carriage. 

CENTER  OF  GRAVITY.  v 

V 

—  1.781 

si  :- 

I 
oo 

1 

V                 1 

fj 

00 

i  ^2 

evi 

t-r— 

V 

T 

-- 

1  yl  ^PROJECTION  OF  AXIS 
7^~l           |       1    OF  CRADLE   TRUNNIONS. 

1 

. 

00                                            to 

Q           00*                                            fe 

06        ^j                               i 

1           1                                   ° 

III 

i 

K 

s. 

m 

Yl 


Area  =  89.59  sq.  ins. 


Ix  =  4923.67  ins.4  Iy  =  455.44  ins.« 

Section  5-5. 
Fig.  58. 


/„  =  5379.11  ins.4 


Forces  Producing  Stress  in  Section  5-5.  —  Neglecting  the 
weight  of  the  pivot  yoke,  most  of  which  is  applied  below  the 
section,  the  forces  acting  are  C/2,  D/Z,  and  E.  The  whole  of 
the  force  E  is  to  be  considered  instead  of  one-half  of  it  because  it 


156  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

is  applied,  through  the  elevating  and  platform  brackets,  to  the 
left  side  of  the  pivot  yoke  only. 

The  force  C/2  =  46647  Ibs.  produces  a  bending  stress  about 
the  axis  XX,  Fig.  58,  a  shearing  stress,  and  a  torsional  stress. 
Its  lever  arm  in  bending,  see  Fig.  22,  is  11.7  ins.  Its  lever  arm 
in  torsion,  see  Fig.  58,  is  .516  in.  obtained  by  considering  the  force 
concentrated  at  the  middle  of  the  trunnion  length  and  measuring 
the  distance  from  its  action  line  to  a  line  perpendicular  to  the  plane 
of  the  section  passing  through  its  center  of  gravity. 

The  force  D/2  =  5359  Ibs.  produces  a  bending  stress  about 
the  axis  XX,  a  stress  of  compression,  and  a  bending  stress  about 
the  axis  YY,  Fig.  58.  Its  lever  arm  with  respect  to  the  axis  XX 
is  3.808  ins.,  and  with  respect  to  the  axis  YY  it  is  .516  in.,  the 
lever  arms  being  determined  by  considering  the  force  concen- 
trated at  the  axis  of  the  trunnions  and  at  the  middle  of  the  trun- 
nion length. 

The  force  E  acts  on  the  left  side  of  the  carriage  only,  being 
applied  to  the  teeth  of  the  elevating  pinion  mounted  on  a  short 
horizontal  shaft  carried  in  bearings  in  the  elevating  bracket. 
The  elevating  bracket  is  bolted  to  the  platform  bracket  and  the 
latter  is  bolted  by  two  upper  and  two  lower  bolts  to  the  rear  face 
of  the  upright  arm  of  the  pivot  yoke.  E  tends  to  rotate  the  plat- 
form bracket  around  the  lower  edge  of  the  surface  of  contact 
between  it  and  the  pivot  yoke.  This  tendency  is  resisted  by  the 
tension  in  the  two  top  bolts  fastening  the  bracket  to  the  pivot 
yoke.  The  lever  arm  of  E  with  respect  to  the  edge  about  which 
the  bracket  tends  to  rotate  being  21.61  ins.  and  that  of  the  tension 
in  the  two  top  bolts  being  20  ins.,  the  intensity  of  the  tension, 
which  will  be  called  Th,  is 

Th  =  4311  X  21.61/20  =  4658  Ibs. 

This  tension  is  transmitted  to  the  pivot  yoke  as  a  horizontal 
force  whose  point  of  application  may  be  taken  as  midway  between 
the  centers  of  the  two  top  bolts  fastening  the  platform  bracket 
to  the  pivot  yoke.  This  point  is  19  ins.  to  the  left  of  the  center 
of  the  carriage  and  3.25  ins.  below  the  axis  of  the  trunnions. 

The  bracket  is  prevented  from  moving  downward  under  the 
action  of  the  vertical  component  of  E  by  the  four  bolts  fasten- 
ing the  bracket  to  the  pivot  yoke.  As  the  shearing  stress  may  be 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES  157 

considered  as  uniformly  distributed  over  the  four  bolts,  the  total 
shearing  stress  in  the  two  top  bolts  is 

4311  cos  14°/2  =  2092  Ibs. 

This  shearing  stress,  which  will  be  called  Tv,  is  transmitted  to 
the  pivot  yoke  as  a  vertical  force,  its  point  of  application  being 
the  same  as  that  of  Th. 

The  two  bottom  bolts  and  the  lower  edge  of  the  surface  of 
contact  between  the  bracket  and  the  pivot  yoke  being  below 
section  5-5,  the  shearing  stress  in  the  bottom  bolts  and  the 
pressure  along  the  contact  edge  must  not  be  considered  in  deter- 
mining the  stresses  in  the  section. 

The  force  Th  =  4658  Ibs.  causes  a  bending  stress  about  the 
axis  XX,  a  shearing  stress,  and  a  torsional  stress.  Its  point  of 
application  being  3.25  ins.  below  the  axis  of  the  trunnions,  its 
lever  arm  in  bending  is  11.7  —  3.25  =  8.45  ins.  As  its  point  of 
application  is  also  19  ins.  to  the  left  of  the  center  of  the  carriage, 
its  lever  arm  in  torsion,  see  Fig.  58,  is  19  -  12.5  -  1.781  =  4.719 
ins.  The  torsional  stress  produced  by  Th  is  opposed  to  that  pro- 
duced by  C/2. 

The  force  Tv  =  2092  Ibs.  causes  a  bending  stress  about  the 
axis  XX,  a  stress  of  compression,  and  a  bending  stress  about 
the  axis  YY.  Its  point  of  application  being  in  the  vertical  plane 
containing  the  rear  face  of  the  upright  arm  of  the  pivot  yoke,  its 
lever  arm  with  respect  to  the  axis  XX,  see  Fig.  58,  is  11.808  ins. 
As  its  point  of  application  is  also  19  ins.  to  the  left  of  the  center 
of  the  carriage,  its  lever  arm  with  respect  to  the  axis  Y  Y  is  19  - 
12.5  -  1.781  =  4.719  ins. 

Bending  Stress  in  Section  5-5  about  XX  as  a  Neutral  Axis.  — 
The  bending  moment  is 

M  =  46647  x  11.7  +  5359  x  3.808  +  4658  x  8.45  +  2092 
x  11.808  =  630239  in.  Ibs. 

and  the  corresponding  bending  stresses  are 

S'"y  =  630239  x  11.808/4923.67  =  1511  Ibs.  per  sq.  in. 
compression  at  the  rear  edge  of  the  section,  and 

S'"y  =  630239  x  11.192X4923.67  =  1433  Ibs.  per  sq.  in. 
tension  at  the  front  edge  of  the  section. 


158  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

Other  Stresses.  —  The  shearing  stress  is 

[46647  +  4658]/89.59  =  573  lbs.  per  sq.  in> 

It  is  apparent  by  inspection  that  the  compressive  stress,  the 


X 

Projection  of  axis  of  trunnions. 

•J52>t 


/-Projection  of 
/  OTi  plane  of  section. 


1_ 


K--J.75— 

CENTER  OF  GRAVITY. 
K- 19.8 

K 3.85J 


LL 


/»!< ;  9.967 


51 


Area  =  81.96  sq.  ins. 


2164.66  ins.4 


186.91  ins.4 


/„  =  2351.57  ins.4 


Section  6-6. 
Fig.  59. 

bending  stress  about  the  axis  YY,  and  the  torsional  stress  are  so 
small  that  it  is  unnecessary  to  compute  them. 

Stresses  in  Section  6-6  of  the  Pivot  Yoke.  —  This  section  is 
indicated  in  Fig.  23  and  shown  in  Fig.  59. 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES  159 

Forces  Producing  Stress  in  Section  6-6.  —  The  force  C/2 
=  46647  Ibs.,  see  Figs.  22  and  23,  produces  in  the  section  a  tor- 
sional  stress,  a  bending  stress  about  the  axis  XX,  and  a  shear- 
ing stress.  Its  lever  arm  in  torsion  is  the  perpendicular  distance 
between  its  action  line  and  the  line  normal  to  the  plane  of  the 
section  passing  through  its  center  of  gravity,  which  distance,  see 
Fig.  59,  is  19.25  -  2.136  =  17.114  ins.  Its  lever  arm  in  bending 
is  the  perpendicular  distance  between  its  action  line  and  the  plane 
of  the  section  equal,  see  Fig.  23,  to  5.75  ins. 

The  force  D/2  =  5359  Ibs.,  see  Figs.  22  and  23,  produces  in 
the  section  a  torsional  stress,  a  bending  stress  about  the  axis 
YY,  and  a  shearing  stress.  Its  lever  arm  in  torsion,  see  Fig.  59, 
is  9.967  -  8.0  =  1.967  ins.  Its  lever  arm  in  bending,  see  Fig.  23, 
is  5.75  ins. 

The  greater  part  of  the  weight  of  the  pivot  yoke  and  shields 
is  applied  above  this  section  so  that  Wpv/2  =  7967  Ibs.  will  be 
considered  as  acting  upon  it.  It  produces  in  the  section  a  tor- 
sional stress,  a  bending  stress  about  the  axis  YY,  and  a  shearing 
stress.  Its  lever  arm  in  torsion,  see  Figs.  23  and  59,  is  .352  in. 
Its  lever  arm  in  bending,  see  Fig.  23,  may  be  taken  as  approxi- 
mately 11.75  ins. 

The  force  E  sin  14°  =  1043  Ibs.  produces  in  the  section  a 
torsional  stress,  a  bending  stress  about  the  axis  XX,  and  a  shear- 
ing stress.  Its  lever  arm  in  torsion,  see  Figs.  22  and  59,  is  19.25 
-  35  sin  14°  -  2.136  =  8.647  ins.  Its  lever  arm  in  bending, 
see  Fig.  23,  is  8.125  ins. 

The  force  E  cos  14°  =  4183  Ibs.  produces  in  the  section  a 
torsional  stress,  a  bending  stress  about  the  axis  YY,  and  a 
shearing  stress.  Its  lever  arm  in  torsion,  see  Fig.  59,  is  35  cos  14° 
+  1.967  =  35.927  ins.  Its  lever  arm  in  bending,  see  Fig.  23,  is 
8.125  his. 

Bending  Stress  in  Section  6-6  about  XX  as  a  Neutral  Axis.  — 
The  bending  moment  is 

46647  x  5.75  -  1043  x  8.125  =  259746  in.  Ibs. 
and  the  corresponding  maximum  bending  stress,  see  Fig.  59,  is 

259746  X  9.967X2164.66  =  1196  Ibs.  per  sq.  in. 
compression  occurring  at  the  surface  CD. 


160  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

Bending  Stress  in  Section  6-6  about  YY  as  a  Neutral  Axis.  — 

The  bending  moment  is 

5359  x  5.75  +  7967  x  11.75  +  4183  x  8.125  =  158413  in.  Ibs. 
and  the  corresponding  maximum  bending  stress,  see  Fig.  59,  is 
158413  x  4.614/186.91  =  3911  Ibs.  per  sq.  in. 

tension  occurring  at  the  surface  AB. 

The  bending  stress  at  the  point  C,  Fig.  59,  due  to  this  bending 
moment  is 

158413  X  2.136/186.91  =  1810  Ibs.  per  sq.  in.  compression. 

Torsional  Stress  in  Section  6-6.  —  The  torsional  moment  is 

46647  X  17.114  +  5359  x  1.967  -  7967  x  .352  -  1043X8.647 
+  4183  X  35.927  =  947318  in.  Ibs. 

and  the  corresponding  maximum  torsional  stress,  which  occurs 
at  C,  is 

947318  V(9.967)2  +  (2.136)2/2351.57  =  4106  Ibs.  per  sq.  in. 

Shearing  Stress  in  Section  6-6.  —  The  shearing  stress  due  to 
the  forces  C/2  and  E  sin  14°  is 

[46647  -  1043]/81.96  =  556  Ibs.  per  sq.  in., 
and  that  due  to  the  forces  D/2,  Wpy/2,  and  E  cos  14°  is 
[5359  +  7967  +  4183]/81.96  =  214  Ibs.  per  sq.  in. 

As  these  two  shearing  stresses  are  due  to  forces  whose  action  lines 
are  perpendicular  to  each  other,  the  resultant  shearing  stress  is 

V(556)2  +  (214)2  =  596  Ibs.  per  sq.  in. 

i 
Maximum  Resultant  Stress  in  Section  6-6.  —  The  maximum 

resultant  stress  in  the  section  occurs  at  C.  It  is  a  compressive 
stress  due  to  combining  the  torsional  stress  at  this  point  with  the 
resultant  compressive  stress  there  caused  by  the  bending  moments 
about  the  axes  XX  and  YY,  respectively.  The  stress  due. to  the 
bending  moment  about  the  axis  XX  being  1196  Ibs.  per  sq.  in. 
compression  and  that  due  to  the  bending  moment  about  YY  being 
1810  Ibs.  per  sq.  in.  compression,  their  resultant  is  1196  +  1810 
=  3006  Ibs.  per  sq.  in.  compression.  Combining  this  with  the 


HWINE 

CANNON 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES  161 

torsional  stress  of  4106  Ibs.  per  sq.  in.  at  C  by  equation  (24)  we 
have 


Stc  =  (3006/2)  +  V(4106)2  +  (3006)2/4  =  5876  Ibs.  per  sq.  in. 

compression  as  the  maximum  resultant  stress  in  the  section. 

The  resultant  shearing  stress  at  C  due  to  the  torsion  and  the 
compression  is,  from  equation  (25), 

S.c  =  V(4106)2  +  (3006)2/4  =  4373  ibs.  per  sq>  in. 

The  shearing  stress  caused  by  the  bending  forces  (but  not  the 
torsional  stress)  in  reality  varies  from  a  maximum  at  the  neutral 
axis  to  zero  at  the  extreme  fiber  and  is,  therefore,  negligible  at  the 
point  C. 

Stresses  in  Section  7-7  of  the  Stem  of  the  Pivot  Yoke.  - 
This  section  is  indicated  in  Fig.  22  and  shown  in  Fig.  60.     Con- 


Area  =  122.52  sq.  ins.  /  =  2726.79  ins.4 

Section  7-7. 
Fig.  60. 

sidering  only  the  forces  applied  below  the  section,  the  force 
H  =  62807  Ibs.,  see  Fig.  22,  produces  bending  and  shear  in  the 
section,  and  the  force  /  =  30833  Ibs.  produces  bending  and  com- 
pression. The  weight  of  the  pivot  yoke  is  not  taken  into  account 
as  it  is  considered  as  applied  above  this  section. 
The  total  bending  moment  is 

62807  x  28  +  30833  x  7  =  1974427  in.  Ibs. 


162 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


and  the  corresponding  maximum  bending  stress  is,  see  Fig.  60, 
1974427  X  8/2726.79  =  5793  Ibs.  per  sq.  in. 

compression  at  the  rear  and  tension  at  the  front  of  the  section. 
The  shearing  stress  due  to  the  force  H  is 

62807/122.52  =  513  Ibs.  per  sq.  in. 
The  compression  due  to  the  force  /  is 

30833/122.52  =  252  Ibs.  per  sq.  in. 

The  maximum  resultant  stress,  therefore,  is  one  of  compression. 
It  occurs  at  the  rear  of  the  section  and  is  equal  to 

5793  +  252  =  6045  Ibs.  per  sq.  in. 

Stresses  in  Section  8-8  of  the  Stem  of  the  Pivot  Yoke.  — 

This  section  is  indicated  in  Fig.  22  and  shown  in  Fig.  61. 


K 12 -H 


Area  =  83.63  sq.  ins.          /  =  949.01  ins.* 

Section  8-8. 
Fig.  61. 

The  force  H,  see  Fig.  22,  produces  bending  and  shear  in  the 
section,  and  the  force  /  produces  bending  and  compression. 
The  total  bending  moment  is 

62807  x  12  +  30833  x  7  =  969515  in.  Ibs. 
and  the  corresponding  maximum  bending  stress  is,  see  Fig.  61, 

969515  X  6/949.01  =  6130  Ibs.  per  sq.  in. 
compression  at  the  rear  and  tension  at  the  front  of  the  section. 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES  163 

The  shearing  stress  due  to  the  force  H  is 

62807X83.63  =  751  Ibs.  per  sq.  in. 

The  compression  due  to  the  force  /  is 

30833X83.63  =  369  Ibs.  per  sq.  in. 

The  maximum  resultant  stress,  therefore,  is  one  of  compression. 
,It  occurs  at  the  rear  of  the  section  and  is  equal  to 

6130  +  369  =  6499  Ibs.  per  sq.  in. 

Stresses  in  Section  9-9  of  the  Stem  of  the  Pivot  Yoke,  — 

This  section  is  indicated  in  Fig.  22  and  shown  in  Fig.  62. 


Area  =  47.71  sq.  ins.  /  =  302.01  ins.4 

Section  9-9. 
Fig.  62. 

The  force  H,  see  Fig.  22,  produces  bending  and  shear  in  the 
section.  The  force  I  which  acts  on  sections  7-7  and  8-8  cannot 
be  considered  as  acting  in  this  case  for  its  point  of  application  is 
above  the  section,  and  only  the  forces  whose  points  of  application 
are  below  it  are  being  considered,  in  accordance  with  the  general 
principles  already  explained. 

The  bending  moment  at  the  section  is 

62807  x  7.875  =  494605  in.  Ibs. 

and  the  corresponding  maximum  bending  stress  is,  see  Fig.  62, 
494605  X  4.5X302.01  =  7370  Ibs.  per  sq.  in. 

compression  at  the  rear  and  tension  at  the  front  of  the  section. 
The  shearing  stress  is 

62807X47.71  =  1316  Ibs.  per  sq.  in. 


164 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


STRESSES  IN  THE  PEDESTAL 

83.  Stresses  in  Section  10-10  of  the  Pedestal.  —  This  sec- 
tion is  indicated  in  Fig.  24  and  shown  in  Fig.  63.  It  is  a  horizon- 
tal section  taken  through  the  centers  of  the  four  hand-holes  in  the 
pedestal.  It  is  assumed  that  the  axis  of  the  gun  when  fired  is 
immediately  above  the  line  joining  the  centers  of  two  of  the  hand- 
holes;  and  the  cover  plates  over  the  holes  are  neglected.  Neg- 


Area  =  134.28  sq.  ins.  /  =  14196  ins.4 

Section  10-10. 
Fig.  63. 

lecting  the  weight  of  the  pedestal  as  the  bulk  of  it  is  below  this 
section,  and  considering  the  forces  whose  points  of  application  are 
above  the  section,  the  only  force  acting,  see  Fig.  24,  is  G  =  155057 
Ibs.    This  produces  bending  and  shear  in  the  section. 
Its  bending  moment  is 

155057  x  16.37  =  2538283  in.  Ibs. 
and  the  corresponding  maximum  bending  stress  is 

2538283  x  15/14196  =  2682  Ibs.  per  sq.  in. 

tension  at  A  and  compression  at  B. 
The  shearing  stress  in  the  section  is 

155057/134.28  =  1155  Ibs.  per  sq.  in. 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES 


165 


Stresses  in  the  Foundation  Bolts.  —  There  are  sixteen  founda- 
tion bolts  fastening  the  pedestal  to  the  gun  platform,  each  of 
which  is  1.75  ins.  in  diameter  at  the  top  of  the  threads  and  1.564 
ins.  in  diameter  at  the  bottom  of  the  threads.  The  positions  of 
these  bolts  are  shown  in  Fig.  64. 


Fig.  64. 

Assume  that  the  holding  down  force  L  =  69441  Ibs.,  see  Fig. 
24,  is  exerted  through  the  five  bolts  included  in  the  quadrant  at 
the  front  of  the  pedestal  as  indicated  in  Fig.  64.  Then,  the  area 
of  each  bolt  at  the  bottom  of  the  threads  being  1.921  sq.  ins.,  the 
stress  in  each  is 

69441/[5  x  1.921]  =  7230  Ibs.  per  sq.  in.  tension. 

Stresses  in  the  Flange  at  the  Rear  of  the  Pedestal.  —  The 

stresses  in  this  part  of  the  flange  are  larger  than  those  which 
occur  at  the  front  owing  to  the  greater  bending  moment  at  the 
rear. 

Although  the  force  T  is  shown  in  Fig.  24  as  acting  at  the  ex- 
treme rear  point  of  the  flange  of  the  pedestal  it  is  in  reality  dis- 
tributed over  the  rear  half  of  the  flange,*  the  intensity  of  the 
pressure  per  unit  of  area  varying  from  a  maximum  at  the  extreme 

*  Part  of  the  force  T  is  also  distributed  over  the  bottom  of  the  pedestal  in- 
side the  flange  but  in  its  vicinity. 


166  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

rear  point  to  zero  at  the  diameter  perpendicular  to  the  axis  of  the 
gun.  Under  the  circumstances,  however,  it  will  be  safe  to  assume 
that  the  force  is  uniformly  distributed  over  the  portion  of  the 
flange  within  the  quadrant  at  the  rear  and  outside  the  dangerous 
section  thereof,  which  occurs  along  the  line  where  the  fillet,  see 
Fig.  24,  joins  the  upper  surface  of  the  flange,  this  line  being  the 
circumference  of  a  circle  47.3  ins.  in  diameter  tangent  to  the 
foundation  bolt  holes  on  the  inside.  The  portion  of  the  flange 
under  consideration  may  be  taken  as  a  cantilever  whose  dangerous 
section  is  TT  x  47.3X4  =  37.15  ins.  wide,  and,  since  the  flange  is 
2  ins.  thick,  the  dangerous  section  has  a  moment  of  inertia  about 
its  neutral  axis  of  24.77  ins.4  and  an  area  of  74.3  sq.  ins.  The 
outer  radius  of  the  flange  being  28.5  ins.  and  the  radius  of  the 
dangerous  section  23.65  ins.,  the  cantilever  can  be  taken  as 
28.5  —  23.65  =  4.85  ins.  long,  and  the  bending  moment  of  the 
force  T  uniformly  distributed  over  it  is,  from  equation  (19), 

103897  x  (4.85)2      oc1ncn  . 

4.85  x  2          =251950  in.  Ibs. 

The  corresponding  maximum  bending  stress  is 

251950  x  1/24.77  =  10172  Ibs.  per  sq.  in. 

compression  at  the  upper  and  tension  at  the  lower  surface  of  the 
flange. 
The  shearing  stress  is 

103897/74.3  =  1398  Ibs.  per  sq.  in. 
The  compression  due  to  the  force  S,  see  Fig.  24,  is 
92250/74.3  =  1242  Ibs.  per  sq.  in. 

The  maximum  resultant  stress,  which  occurs  at  the  upper  edge 
of  the  section  is,  therefore, 

10172  +  1242  =  11414  Ibs.  per  sq.  in.  compression. 

STRESSES  IN  PAETS  OF  THE  6-INCH  DISAPPEARING 
CARRIAGE,  MODEL  OF  1905  Ml. 

84.  Stresses  in  Section  11-11  of  (he  Gun  Levers.  —  The  gun- 
lever  axle  and  the  part  of  the  gun  levers  above  it  are  shown 
in  Fig.  65,  in  which  are  also  indicated  the  forces  (in  full  lines) 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES 


167 


168 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


exerted  on  the  levers  by  the  gun  and  the  position  of  section  11-11. 
The  values  of  the  forces  P  and  P\  were  determined  in  Chapter  III, 
see  equations  (!'")  and  (2'"),  page  89,  to  be  179338  Ibs.  and 
92902  Ibs.,  respectively. 
Section  11-11  is  shown  in  Fig.  66. 


1 
1 
t 

1 

-1.5 

T 
i 

*  j 

1 

i 
ITS                           Area  =  94.17  sq.  ins. 
&                                /  =  3882  ins.4 
06 
1 

i 

-X 


1± 


Section  11-11. 
Fig.  66. 

Method  of  Computing  Stresses  in  Parts  not  in  Equilibrium.  — 

The  stresses  in  this  section  cannot  be  obtained  in  the  same  way 
as  were  the  stresses  in  the  parts  of  the  3-inch  field  carriage  and 
the  5-inch  barbette  carriage  because  those  parts  were  in  equi- 
librium under  the  forces  acting  upon  them;  while  the  gun  levers 
have  linear  accelerations  in  the  vertical  and  horizontal  directions 
and  an  angular  acceleration  about  their  center  of  mass,  and  a 
part  of  each  of  the  forces  acting  on  them  is  taken  up  in  producing 
these  accelerations  without  causing  internal  stresses  in  the  parts. 
If,  however,  a  section  such  as  11-11,  Fig.  65,  be  taken  through  any 
part  of  a  body  not  in  equilibrium,  the  internal  stresses  acting  in 
that  section  must  be  such  that,  in  connection  with  the  external 
forces  acting  on  the  part  of  the  body  on  one  side  (either  side)  of 
the  section,  they  will  produce  in  that  part  the  accelerations  of 
translation  and  rotation  possessed  by  it.  Therefore,  considering 
the  part  of  the  gun  levers  above  section  11-11,  we  may  form  three 
equations  between  the  known  external  forces  and  the  unknown 
stresses  in  section  11-11  stating:  (a)  that  the  sum  of  the  compo- 
nents of  both  forces  and  stresses  in  a  vertical  direction  is  equal  to 
the  mass  of  the  part  of  the  gun  levers  above  section  11-11  multi- 
plied by  its  acceleration  in  that  direction;  (6)  that  the  sum  of  the 
components  in  the  horizontal  direction  is  equal  to  the  mass  multi- 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES  169 

plied  by  its  acceleration  in  the  horizontal  direction;  and  (c)  that 
the  sum  of  the  moments  of  both  forces  and  stresses  with  respect  to 
the  center  of  mass  of  the  part  of  the  gun  levers  above  section  11-11 
is  equal  to  the  moment  of  inertia  of  that  part,  with  respect  to  an 
axis  passing  through  its  center  of  mass  perpendicular  to  the  plane 
of  the  forces,  multiplied  by  the  angular  acceleration  of  the  part 
about  that  axis.  If  these  three  equations  do  not  contain  more 
than  three  unknown  quantities,  the  internal  stresses  in  section 
11-11  can  be  determined. 

Resolution  of  the  Total  Stresses  in  Section  11-11  into  Un- 
known Horizontal  and  Vertical  Components  Ps  and  Pt  and  an 
Unknown  Couple  YY.  —  Referring  to  Fig.  65,  the  total  stresses 
in  section  11-11  can  be  resolved  into  unknown  horizontal  and 
vertical  components  P8  and  Pt,  respectively,  and  an  unknown 
couple  yy  as  shown  by  the  dot  and  dash  force  lines,  these  being 
the  most  general  assumptions  that  can  be  made  with  regard  to 
any  set  of  unknown  co-planar  forces  or  stresses  acting  there.  The 
point  of  application  of  P,  and  Pt  can  be  taken  at  the  center  of  the 
section.  They  will  be  assumed  to  act  in  opposite  directions  to 
the  external  forces  P  and  PI,  and  if  this  assumption  is  not  correct 
it  will  be  shown  by  a  negative  sign  opposite  the  numerical  value 
of  one  or  both  of  the  stresses  obtained  from  the  equations.  The 
assumption  of  wrong  directions  for  the  component  stresses  will 
not  affect  the  correctness  of  the  numerical  results.  The  sum  of 
the  components  of  Pa  and  Pt  parallel  to  section  11-11  is  the  total 
shearing  stress  in  the  section,  and  the  sum  of  their  components 
normal  to  the  section  is  the  total  stress  of  tension  or  compression 
therein.  Since  Pt  and  Pt  are  applied  at  the  center  of  the  section 
they  can  have  no  tendency  to  rotate  it  and  will,  therefore,  have  no 
effect  on  the  bending  moment  at  the  section.  The  couple  YY  can 
produce  neither  shear  nor  simple  tension  or  compression  in  the  sec- 
tion since  the  individual  forces  of  a  couple  are  equal  in  intensity 
but  opposite  in  direction.  It  does,  however,  tend  to  rotate  the  sec- 
tion about  the  neutral  axis  XX,  Fig.  66,  and  it  is,  therefore,  the 
bending  moment  at  the  section. 

Determination  of  the  Values  of  P.,  Pt,  and  YY.  —  Writing  the 
three  equations  expressing  the  relations  between  the  known 
forces,  P  =  179338  Ibs.  and  Pi  =  92902  Ibs.,  and  the  unknown 
total  stresses  in  section  11-11,  and  noting  that  as  yy  is  a  couple 


170  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

it  need  only  appear  in  the  equation  of  moments,  we  have,  since 
M  =  40  and  Smr2  =  230,  see  Fig.  65, 

179338  -  P,  =  40  d*x/dt*  (41) 

92902  -  Pt  =  40  d*y/dt2  (42) 

179338  x  22c°sll°  f28'75  C°S 


12  "I  12 

22  sin  11°       „       [28.75  sin  11°  +  3.682  cos  13 

—  Jrt  X 


12  "(  12 

-  yy  =  230  d^/dt2  (43) 

d^x/dfi  and  d^y/dP  being  the  accelerations  of  the  center  of 
mass  of  the  part  of  the  gun  levers  above  section  11-11  in  the 
horizontal  and  vertical  directions,  respectively,  and  d?<t>/dP 
being  the  angular  acceleration  of  that  part  about  the  axis  through 
its  center  of  mass  perpendicular  to  the  plane  of  the  forces.  Since 
the  unit  of  force  is  the  pound,  the  linear  accelerations  must  be 
expressed  in  feet  per  second  per  second  and  the  moment  of  inertia 
in  mass  units  and  feet. 

The  horizontal  acceleration  in  equation  (41)  may  be  obtained 
from  equation  (43),  page  82,  by  neglecting  the  term  containing 
(d<j>/dt}z  as  being  too  small  to  be  considered;  by  substituting  for 
d2<^/dt2  its  value  of  224.98  radians  per  sec.  per  sec.  from  equation 
(7/F),  page  89;  by  substituting  for  a,  <f>,  and  (0  —  0)  their  values 
of  4.3333  ft.,  13°,  and  11°,*  respectively,  from  table  3,  page  85; 
and  by  replacing  c,  the  distance  of  the  axis  of  the  gun  trunnions 
from  the  axis  of  the  trunnions  of  the  gun-lever  axle,  by  40/12  ft., 
the  distance  of  the  center  of  mass  of  the  part  of  the  gun-levers 
above  section  11-11  from  the  axis  of  the  trunnions  of  the  gun- 
lever  axle;  whence 

dzx/dt*  =  (4.3333  cos  13°  +  ^cos  11°)  224.1 

=  1679  ft.  per  sec.  per  sec. 
and 

M  d*x/dt*  =  40  x  1679.5  =  67180. 


See  article  56,  page  86. 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES  171 

By  making  the  same  omission  and  substitutions  in  equation  (48), 
page  83,  and  substituting  for  a  its  value  of  1°  20'  from  table  3,  we 
obtain  as  the  vertical  acceleration  in  equation  (42) 

(Py/dP  =  (4.3333  tan  1°  20'  cos  13°  -  ^sin  11°)  224.08 

=  —120.5  ft.  per  sec.  per  sec. 
and  Md*y/dt2  =  40  (-120.5)  =  -4820 

The  angular  acceleration  of  the  part  of  the  gun  levers  above 
section  11-11  about  an  axis  passing  through  its  center  of  mass 
perpendicular  to  the  plane  of  the  forces  is  the  same  as  that  of  the 
gun  levers,  complete,  about  the  axis  of  the  gun-lever  pins,  and  is 
224.08  radians  per  sec.  per  sec.,  whence 

Swr2  x  ffij/dP  =  230  x  224.08  =  51543 

Substituting  these  values  of  Md?x/dP,  Md*y/dP,  and 
2rar*  d*<t>/dt2  in  equations  (41),  (42),  and  (43),  respectively,  the 
only  unknown  quantities  remaining  therein  are  P,,  Pt,  and  YY 
which  may  be  determined  to  have  the  following  values: 

P,  =  112158  Ibs. 

p  —  97722  Ibs. 

YY  =  420839'ft.  Ibs.  =  5050068  in.  Ibs. 

Shearing  Stress  in  Section  11-11.  —  The  total  shearing  stress 
in  section  11-11  is  the  algebraic  sum  of  the  components  of  P,  and 
Pt  parallel  to  the  section.  It  is  equal  to 

112158  cos  13°  -  97922  sin  13°  =  87300  Ibs. 
and  the  shearing  stress  per  unit  of  area  is,  see  Fig.  66, 
87300/94.17  =  927  Ibs.  per  sq.  in. 

Tensile  Stress  in  Section  11-11.  —  The  total  longitudinal 
stress  in  this  section  is  the  sum  of  the  components  of  Ps  and  Pt 
perpendicular  to  the  section;  and,  since  these  components  act 
away  from  the  section,  the  stress  is  one  of  tension.  It  is  equal  to 

112158  sin  13°  +  97922  cos  13°  =  120446  Ibs. 
and  the  tensile  stress  per  unit  of  area  is 

120446/94.17  =  1279  Ibs.  per  sq.  in. 


172  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

Bending  Stress  in  Section  11-11.  —  The  bending  moment 
YY  is  5050068  in.  Ibs.  and  the  corresponding  maximum  bending 
stress  is,  see  Fig.  66, 

5050068  x  8.575/3882  =  11155  Ibs.  per  sq.  in. 

tension  at  the  front  and  compression  at  the  rear  of  the  section. 

Maximum  Resultant  Stress  in  Section  11-11.  —  The  maximum 
resultant  stress  in  the  section  is  one  of  tension;  it  occurs  at  the 
front  of  the  section  and  is  equal  to 

1279  +  11155  =  12434  Ibs.  per  sq.  in. 

85.   Stresses  in  the  Trunnions  of  the  Gun-Lever  Axle.  —  The 

gun  levers,  see  Fig.  65,  are  shrunk  and  bolted  to  the  gun-lever 
axle,  and  the  whole  may  be  regarded  as  a  beam  carried  in  bear- 
ings in  the  top  carriage  and  subjected  to  bending  forces  applied 
to  each  end  of  the  gun  levers,  as  shown  in  Fig.  31,  Chapter  III. 
Because  of  the  great  stiffness  of  the  beam  formed  by  the  gun  levers 
and  the  gun-lever  axle  in  connection  with  the  fact  that  the  trun- 
nions are  .01  in.  smaller  in  diameter  than  their  bearings  in  the  top 
carriage,  the  levers  and  axle  may,  like  the  cradle  of  the  5-inch 
barbette  carriage,  model  of  1903,  be  considered  as  a  beam  that  is 
merely  supported  at  its  ends. 

The  forces  acting  on  each  trunnion,  see  equations  (4"')  and 
(5"'),  page  89,  are  the  horizontal  force  P5/2  =  39111.5  Ibs.  and 
the  vertical  force  P4/2  =  67874  Ibs.  The  resultant  of  these 
forces  is 

V(39111.5)2  +  (67S74)2  =  78328  Ibs. 

and  it  is  uniformly  distributed  over  the  surface  of  contact  between 
the  trunnion  and  its  bed  in  the  top  carriage.  This  surface  is  10 
ins.  long,  being  .26  in.  shorter  than  the  trunnion  to  provide  a 
clearance  of  .01  in.  on  the  inside  and  to  allow  the  trunnion  to  pro- 
ject through  for  a  distance  of  .25  in.  on  the  outside. 

The  bending  moment  at  section  AB,  Fig.  65,  is  greater  than 
at  any  other  section  of  the  trunnion.  It  may  be  obtained  by 
considering  the  resultant  force  concentrated  at  the  middle  of  the 
length  of  the  surface  of  contact  between  the  trunnion  and  its  bed, 
whence 

M  =  78328  x  5.01  =  392423  in.  Ibs. 


STRESSES  IN  PARTS  OF  GUN  CARRIAGES  173 

The  diameter  of  the  trunnion  being  8.99  ins.,  the  area  of  any  right 
section  thereof  is  63.48  sq.  ins.  and  the  moment  of  inertia  of  the 
section  with  respect  to  its  neutral  axis  is  320.72  ins.4  The  maxi- 
mum bending  stress  in  the  section  is,  therefore, 

392423  x  4.495/320.72  =  5500  Ibs.  per  sq.  in. 

tension  at  the  rear  and  compression  at  the  front  of  the  section. 
The  shearing  stress  in  section  AB  is 

78328/63.48  =  1234  Ibs.  per  sq.  in. 

86.  Stresses  in  the  Elevating  Arm  and  Other  Parts  of  the  6- 
inch  Disappearing  Carriage,  Model  of  1905  Ml.  —  The  stresses 
in  any  section  of  the  elevating  arm,  which  rotates  about  an  axis 
on  the  rear  transom  of  the  carriage,  must  be  determined  by  the 
same  method  as  the  stresses  in  section  11-11  of  the  gun  levers. 
The  method  of  computing  the  stresses  in  the  remaining  parts  of 
the  carriage  is  the  same  in  principle  as  that  followed  throughout 
this  text  in  the  determination  of  the  stresses  in  parts  of  the  3-inch 
field  carriage  and  of  the  5-inch  barbette  carriage. 


CHAPTER  V. 
TOOTHED  GEARING. 

87.  Definition.  —  Ratio  of  Angular  Velocities.  —  In  order  to 
transmit  rotary  motion  from  one  shaft  to  another  of  a  gun  car- 
riage, or  of  any  other  machine,  toothed  gearing  is  employed. 
This  consists  of  a  toothed  wheel  fastened  to  one  shaft  which  when 
the  shaft  rotates  moves  with  it,  and  by  the  engagement  of  its 
teeth  with  those  of  a  second  wheel  fastened  to  a  second  shaft 
causes  a  rotation  of  the  second  wheel  and  shaft. 


Fig.  67. 

Let  a  and  6,  Fig.  67,  be  two  wheels  mounted  on  parallel  shafts 
A  and  B  and  pressed  together  with  considerable  force  so  that 
the  friction  at  their  line  of  contact  projected  on  P  is  sufficient  to 
make  either  wheel  turn  if  the  other  is  turned.  Suppose  the  shaft 
B  be  rotated  counter-clockwise  as  shown  by  the  arrow.  Then  b 
will  rotate  in  the  ^ame  direction  and  because  of  the  friction  at  P 
the  wheel  a  and  the  shaft  A  will  be  forced  to  rotate  in  the  opposite 
direction  34  shown  by  the  arrow  on  a. 

Let     V   be  the  surface  velocity  of  the  wheel  6, 
n   the  radius  of  the  wheel  6, 
W6  the  angular  velocity  of  the  wheel  6, 

174 


TOOTHED  GEARING  175 

Nb  the  number  of  revolutions  per  minute  of  the  wheel  6, 

ra    the  radius  of  the  wheel  a, 

(j)a  the  angular  velocity  of  the  wheel  a,  and 

Na  the  number  of  revolutions  per  minute  of  the  wheel  a. 

Assuming  that  there  is  no  slipping  at  the  line  of  contact,  the 
surface  velocities  of  a  and  b  must  be  the  same  and,  consequently, 

V=  rv&b  =  fa(n)a 
or  cua/co6  =  rb/ra 

That  is,  the  angular  velocities  of  the  two  wheels  are  inversely  as  their 
radii. 

Since  the  number  of  revolutions  per  minute  of  a  wheel  varies 
directly  as  its  angular  velocity  we  may  write 

Na/Nb  =  ua/ub  =  rb/ra 

or  the  numbers  of  revolutions  per  minute  of  the  two  wheels  are  in- 
versely as  their  radii. 


Fig.  68. 

88.  Necessity  for  Teeth.  —  Tooth  Surfaces  Must  Have  Cer- 
tain Definite  Forms.  —  The  wheels  a  and  6  are  called  friction 
wheels  and  are  used  to  a  limited  extent  in  machinery  to  transmit 
rotary  motion  from  one  shaft  to  another.  If  any  great  amount 
of  power  has  to  be  transmitted,  however,  the  wheels  will  slip  as 
the  greatest  force  that  can  be  transmitted  between  their  surfaces 
is  equal  to  the  normal  pressure  there  multiplied  by  the  coefficient 
of  friction.  To  prevent  slipping  the  surface  of  each  wheel  is 
provided  with  projections  called  teeth  which  engage  with  corre- 
sponding teeth  on  the  other  wheel  as  shown  in  Fig.  68. 


176 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


It  is  evident  whatever  be  the  shape  of  the  teeth,  provided  they 
are  of  the  same  uniform  size  and  uniformly  spaced  on  both  wheels 
and  the  recesses  between  them  are  large  enough  for  them  to  enter, 
that  the  numbers  of  revolutions  per  minute  wil  be  inversely 
proportional  to  the  radii  of  the  wheels  or  to  the  numbers  of  the 
teeth,  since  these  must  vary  directly  as  the  radii;  and  it  may  not, 
therefore,  be  apparent  at  first  why  the  shape  of  the  teeth  should 
not  be  arbitrarily  selected.  It  is  necessary,  however,  that  the 
machinery  shall  run  smoothly,  without  jerks,  and  in  most  cases 
it  is  very  essential  that  the  velocity  of  one  moving  part  of  a  ma- 
chine shall  at  all  times  have  definite  relations  to  the  velocities  of 
the  other  moving  parts.  Therefore,  not  only  must  the  numbers 
of  revolutions  per  minute  of  the  wheels  have  a  definite  relation 
to  each  other  but  the  angular  velocities  at  each  instant  of  time 
must  have  this  same  definite  ratio.  This  is  possible  only  when 
the  tooth  surfaces  have  certain  definite  forms. 


^  ^CIRCULAR  PITCH.-*!  |       ^ 


Cu 

Fig.  69. 


89.  Outline  of  Gear-Teeth.  —  Pitch  Circumference.  —  Cir- 
cular Pitch.  —  Diametral  Pitch.  —  Fig.  69  shows  the  outline  of 
several  gear-teeth. 

The  pitch  circumference  *  shown  in  the  figure  is  the  projection 
of  the  imaginary  cylindrical  surface  which,  rolling  on  a  similar 


*  The  circumferences  of  the  pitch,  addendum,  and  root  circles  are  generally 
referred  to  as  circles,  not  circumferences. 


TOOTHED  GEARING  177 

cylindrical  surface  of  another  gear-wheel,  would  cause  the  two 
wheels  to  rotate  together  with  the  same  relative  angular  velocities 
as  would  be  caused  by  the  engagement  of  the  teeth  of  the  wheels; 
or  the  pitch  circumference  may  be  considered  as  the  projection 
of  the  cylindrical  surface  of  a  friction  wheel  such  as  shown  in  Fig. 
67  on  which  teeth  have  been  fastened  as  shown  in  Fig.  68  to  make 
the  motion  more  reliable,  it  being  understood  that  the  ratio  be- 
tween the  angular  velocities  of  the  wheels  in  both  figures  is  the 
same. 

The  circle  drawn  through  the  outer  ends  of  the  teeth  is  called 
the  addendum  circle  and  that  drawn  through  their  inner  ends  is 
called  the  root  circle.  The  distance  between  the  pitch  and  adden- 
dum circumferences  measured  on  a  radial  line  is  called  the  adden- 
dum and  that  between  the  pitch  and  root  circumferences  measured 
on  a  radial  line  is  called  the  root.  The  term  addendum  is  also 
frequently  used  to  designate  the  part  of  the  tooth  outside  the 
pitch  surface,  and  the  term  root  to  designate  the  part  of  the  tooth 
inside  that  surface.  The  working  surface  of  the  portion  of  the 
tooth  outside  the  pitch  surface  is  called  the  face  of  the  tooth  and 
the  working  surface  inside  the  pitch  surface  is  called  the  flank  of 
the  tooth.  The  distance  between  corresponding  points  of  two 
adjacent  teeth  measured  on  the  pitch  circumference  is  called  the 
circular  pitch.  The  diametral  pitch  is  the  number  of  teeth  per 
inch  of  diameter  of  the  pitch  circle;  thus  if  pi  represent  the  diam- 
etral pitch,  the  number  of  teeth  of  a  gear-wheel  whose  pitch 
diameter  is  d  is  pid.  From  the  definition  of  circular  pitch  the 
number  of  teeth  of  the  same  wheel  expressed  in  terms  of  the  circu- 
lar pitch,  which  will  be  represented  by  p,  is  vd/p.  Since  pid 
=  ird/p  we  have  pi  =  n/p,  or  the  diametral  pitch  is  equal  to  TT 
divided  by  the  circular  pitch.  The  diametral  pitch  is  the  one 
more  frequently  used  to  express  the  pitch  of  the  teeth  of  gear- 
wheels. 

90.  Angular  Velocity  of  Rotating  Wheel.  —  Let  the  wheel  a 
shown  in  Fig.  70  be  rotating  about  its  center  0  and  let  the  velocity 
of  any  point  b  on  its  circumference  be  represented  in  magnitude 
and  direction  by  the  line  marked  V.  Then  if  r  be  the  radius  of 
the  wheel  its  angular  velocity  w  will  be  V ' /r.  Resolve  the  velocity 
V  into  two  components  normal  to  each  other,  Vn  and  V'n  as  shown.: 
Let  a  be  the  angle  between  the  directions  of  V  and  Vn  and  let  rn 


178 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


be  the  perpendicular  distance  from  the  center  of  the  wheel  to  the 
direction  line  of  Vn.    Then 


and 
whence 
and,  similarly 


Vn  =  V  COS  a 

rn  =  r  cos  a 


=      = 


rn       r  cos  a       r 

V'nT'n    =   CO. 


Fig.  70. 

If  the  wheel  be  increased  in  size  so  that  the  point  b  no  longer  lies 
on  its  circumference,  the  relations  just  established  will  still  hold 
for  the  velocity  at  the  point  6,  r  in  this  case,  however,  will  be  the 
perpendicular  distance  from  the  center  of  the  wheel  to  the  direc- 
tion line  of  V;  whence  we  may  write 

//  the  velocity  of  any  point  of  a  wheel  rotating  about  its  center 
be  resolved  into  two  components  normal  to  each  other,  the  angular 
velocity  of  the  wheel  will  be  equal  to  either  component  divided  by 
the  perpendicular  distance  from  the  center  of  the  wheel  to  the 
direction  line  of  that  component. 


TOOTHED  GEARING 


179 


91.   Condition  to  be  Fulfilled  by  Tooth  Curves.  —  Let  Ta  and 

Tb,  Fig.  71,  be  two  teeth  in  contact  at  the  point  m,  and  C  and  C\ 
be  the  centers  of  the  wheels  to  which  Ta  and  Tb  belong.  Since 
the  tooth  curves  are  tangent  at  m  they  have  at  this  point  a  com- 
mon normal  NNi  which  intersects  the  line  of  centers  Cd  at  some 
point  P.  Suppose  the  wheel  B  to  be  rotating  counter-clockwise 


Fig.  71. 

around  its  center  Ci  driving  the  wheel  A  around  its  center  C  in  a 
clockwise  direction;  then  the  velocity  of  the  point  m  on  the  tooth 
Tb  may  be  represented  by  the  line  mb  drawn  tangent  to  the  circle 
described  through  m  about  the  center  Ci,  and,  similarly,  the 
velocity  of  the  point  m  on  the  tooth  Ta  may  be  represented  by  the 
line  ma  drawn  tangent  to  the  circle  described  through  m  about 
the  center  C.  Let  the  point  m  on  the  tooth  Tb  be  called  mb  and 
the  point  m  on  the  tooth  Ta  be  called  ma,  and  resolve  the  velocities 
of  mb  and  ma  into  components  parallel  and  perpendicular  to  the 
common  normal  NN\. 

While  the  velocities  of  mb  and  ma  vary  in  direction  and  in  mag- 
nitude, the  components  of  these  velocities  in  the  direction  of  their 
common  normal  must  be  equal  during  the  time  mb  and  ma  are  in 
contact,  for  otherwise  the  teeth  would  separate  or  one  surface 
would  penetrate  the  other.  Representing  by  vn  the  equal  com- 
ponent volecities  of  mb  and  ma  in  the  direction  of  the  common 


180 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


normal,  and  dropping  the  perpendiculars  CiNi  and  CN  on  this 
normal,  the  angular  velocities  «&  and  coa  of  the  wheels  B  and  A, 
respectively,  are  (see  article  90) 

and  coa  =  vn/CN 
i  =  CP/dP 


whence  co6/coa 

or 

The  angular  velocities  of  the  two  wheels  are  inversely  as  the 
segments  into  which  the  line  of  centers  is  cut  by  the  common 
normal  to  the  tooth  curves  at  their  point  of  contact. 

The  condition  to  be  fulfilled  by  tooth  curves  in  order  that  the 
ratio  of  the  angular  velocities  of  the  wheels  shall  be  constant  at 
every  instant  is,  therefore,  that  the  common  normal  to  the  tooth 
curves  at  their  point  of  contact  shall  always  pass  through  a  fixed  point 
on  the  line  of  centers. 

In  order  that  the  wheels  shall  have  the  same  angular  velocities 
as  their  pitch  circles  the  common  normal  to  the  tooth  curves  at  their 
point  of  contact  must  intersect  the  line  of  centers  at  the  pitch  point, 
which  is  the  point  of  tangency  of  the  two  pitch  circles.  Any  number 
of  curves  will  fulfill  this  condition  but  the  two  generally  adopted 
for  tooth  outlines  are  the  involute  curve  and  the  cycloidal  curve. 


B 


92.  The  Involute  Curve.  —  If  a  cord  be  wound  around  a  circle 
and  then  unwound,  keeping  the  unwound  portion  straight,  a  point 
of  the  cord  will  describe  an  involute  to  the  circle.  Let  EC,  Fig. 
72,  be  a  portion  of  a  cord  unwound  from  the  circle  A.  The  point 


ENGINEERING  BUREAU 

CANNON 


TOOTHED  GEARING  181 

C  of  the  cord  before  it  was  unwound  was  at  the  point  Ci  of  the 
circle,  and  the  curve  CJ2C  is  the  involute  to  the  circle  traced  by 
the  point  C.  At  each  instant  C  is  rotating  about  the  point  of 
tangency  of  the  cord  to  the  circle  and,  consequently,  its  direction 
at  that  instant  is  at  right  angles  to  the  tangent  line  drawn  from  it 
to  the  circle.  Any  normal  to  the  involute  is,  therefore,  tangent 
to  the  circle. 

The  Involute  System  of  Gear-Teeth.  —  Let  APB  and  CPD, 
Fig.  73,  be  the  pitch  circles  of  two  wheels  on  which  it  is  desired 
to  construct  teeth  of  such  profile  that  the  ratio  of  their  angular 
velocities  when  running  in  gear  shall  at  each  instant  be  exactly  the 
same  as  the  ratio  of  the  angular  velocities  of  the  pitch  circles  when 
running  together  without  teeth.  Through  the  pitch  point  P  draw 
a  line  NNi  at  an  angle  a  with  the  common  tangent  to  the  pitch 
circles,  and  draw  circles  EF  and  GH  tangent  to  NNi  about  the 
centers  0  and  Oi,  respectively,  of  the  pitch  circles.  Let  the  pro- 
files of  the  teeth  on  APB  be  involutes  to  the  circle  EF  and  the 
profiles  of  those  on  CPD  be  involutes  to  the  circle  GH.  Now 
let  APB  be  rotated  hi  the  direction  of  the  arrow  until  one  of  its 
teeth  is  brought  in  contact  with  a  tooth  on  CPD.  At  the  point 
of  contact  the  curves  on  the  two  teeth  will  have  a  common  normal 
which  must  be  tangent  to  the  circle  EF  because  the  profile  of  the 
tooth  on  APB  is  an  involute  to  that  circle,  and  which  must  also 
be  tangent  to  the  circle  GH  because  the  profile  of  the  tooth  on 
CPD  is  an  involute  to  the  circle  GH.  From  the  figure  it  is  evident 
that  NNi  is  the  only  line  tangent  to  the  circles  EF  and  GH  that 
can  pass  through  the  point  of  contact  of  the  teeth,  and  this  line 
must,  therefore,  be  the  common  normal  to  the  tooth  curves  at 
their  point  of  contact.  If,  however,  the  wheel  APB  be  rotated 
in  the  opposite  direction  the  point  of  contact  will  change  to  the 
other  sides  of  the  teeth  and  the  common  normal  to  the  tooth 
curves  at  their  point  of  contact  will  be  the  line  N'N\,  which  is 
also  a  common  tangent  to  the  circles  EF  and  GH  passing  through 
the  pitch  point  P.  The  point  of  contact  of  the  teeth  must,  there- 
fore, always  lie  on  one  or  the  other  of  the  lines  NNi  and  N'N'i, 
both  of  which  are  common  normals  to  the  tooth  curves  passing 
through  the  pitch  point.  It  is  evident,  therefore,  that  the  in- 
volute to  a  circle  fulfills  strictly  the  condition  required  for  tooth 
profiles  of  gear-wheels. 


182 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


Fig.  73 


TOOTHED  GEARING 


183 


The  circles  EF  and  GH  are  called  base  circles.  Since  the  point 
of  contact  of  the  teeth  is  always  on  one  or  the  other  of  the  lines 
NNi  and  N'N\,  these  lines  are  called  the  lines  of  action.  Neg- 
lecting friction  the  line  of  action  is  the  direction  of  the  pressure 
between  the  teeth.  The  angle  a  between  the  line  of  action  and 
the  common  tangent  to  the  pitch  circles  is  called  the  angle  of 
obliquity.  In  order  that  gears  with  involute  teeth  shall  be  inter- 
changeable they  must  have  the  same  circular  pitch  and  the  same 
angle  of  obliquity.  The  standard  angle  of  obliquity  is  15°. 

The  arc  of  the  pitch  circle  that  subtends  the  angle  through 
which  a  wheel  rotates  from  the  time  when  one  of  its  teeth  comes 
in  contact  with  a  tooth  of  the  other  wheel  until  the  point  of  con- 


Fig.  74. 

tact  reaches  the  pitch  point  P  is  called  the  arc  of  approach;  and 
the  arc  of  the  pitch  circle  that  subtends  the  angle  through  which 
the  wheel  rotates  from  the  time  the  point  of  contact  leaves  the 
pitch  point  until  the  teeth  separate  is  called  the  arc  of  recess. 
The  sum  of  these  arcs  is  called  the  arc  of  action.  In  order  that 
there  shall  always  be  contact  between  the  teeth,  the  arc  of  action 
should  be  at  least  equal  to  the  circular  pitch.  If  it  is  desired  that 
there  shall  always  be  contact  between  two  pairs  of  teeth,  the  arc 
of  action  should,  be  at  least  equal  to  twice  the  circular  pitch. 
The  arc  of  action  may  be  increased  by  lengthening  the  teeth 
up  to  a  certain  limit.  Fig.  74  shows  the  profile  of  the  involute 
tooth. 


184 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


93.  The  Cycloid.  —  Epicycloid.  —  Hypocycloid.  —  If  a  circle 
be  rolled  upon  a  straight  line  a  point  in  its  circumference  will 
trace  a  curve  called  a  cycloid.  If  the  circle  be  rolled  upon  the 

CYCLOID. 


CPICVCLQ/O 


HYPOCYCLOID- 
Fig.  75. 

convex  side  of  another  circle  a  point  in  the  circumference  of  the 
former  circle  will  trace  a  curve  called  an  epicycloid;  and  if  rolled 
upon  the  concave  side  of  the  circle  the  curve  traced  is  called  a 
hypocycloid.  These  curves  are  shown  in  Fig.  75. 


TOOTHED  GEARING 


185 


Let  G  be  the  point  on  the  generating  circle  that  traces  each  of 
the  three  curves  shown  in  Fig.  75.  Then,  since  the  motion  of 
the  generating  circle  and  the  point  G  is  at  each  instant  a  rotation 
about  the  point  of  tangency  P  of  the  generating  circle  and  the 
line  upon  which  it  rolls,  the  direction  of  motion  of  G  is  at  right 
angles  to  the  line  joining  it  to  P,  and  this  line  GP  is  a  normal  to 
the  cycloid,  epicycloid,  or  hypocycloid  as  the  case  may  be. 


D 


The  Cycloidal  System  of  Gear-Teeth.  —  Let  APB  and  CPD, 

Fig.  76,  be  the  pitch  circles  of  two  wheels  on  which  it  is  desired  to 
construct  teeth  of  such  profile  that  the  ratio  of  their  angular 
velocities  when  running  in  gear  shall  at  each  instant  be  exactly 
the  same  as  that  of  the  pitch  circles  when  running  together  with- 
out teeth.  Suppose  the  generating  circle  GPE  be  rolled  on  the 
outside  of  APB  and  that  the  point  G  traces  the  epicycloid  FGH ; 
suppose  also  that  the  generating  circle  be  rolled  on  the  inside  of 
CPD,  the  point  G  tracing  the  hypocycloid  IGK.  Let  the  portion 


186  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

FG  of  the  epicycloid  be  taken  as  the  profile  of  the  part  of  a  tooth 
on  APB  outside  the  pitch  circle,  that  is,  as  the  profile  of  the  face 
of  a  tooth  on  APB  (see  Fig.  69),  and  the  portion  IG  of  the  hypo- 
cycloid  be  taken  as  the  profile  of  the  flank  of  a  tooth  on  CPD; 
and  let  FG  and  IG  be  brought  into  contact  as  shown.  Then  the 
common  normal  to  the  tooth  curves  at  their  point  of  contact  must 
pass  through  the  point  at  which  the  generating  circle  was  tangent 
to  the  pitch  circle  APB  at  the  instant  the  point  G  of  the  epicycloid 
was  being  traced,  and  it  must  also  pass  through  the  point  at  which 
the  generating  circle  was  tangent  to  the  pitch  circle  CPD  at  the 
instant  the  point  G  of  the  hypocycloid  was  being  traced;  and 
further,  since  the  point  of  contact  G  is  a  point  of  both  the  epicy- 
cloid and  the  hypocycloid,  the  generating  circle  must  have  been 
tangent  to  both  of  the  pitch  circles  at  the  instant  the  point  G 
was  being  traced.  The  common  normal  to  the  tooth  curves  at 
their  point  of  contact  must,  therefore,  pass  through  the  pitch 
point  P.  As  this  is  true  for  any  point  of  contact  of  the  tooth 
curves  FG  and  IG,  it  follows  that  when  the  profile  of  the  face  of  a 
tooth  on  one  gear  is  part  of  the  epicycloid  traced  by  a  point  of  the 
generating  circle  rolling  on  the  outside  of  the  pitch  circle  of  that 
gear,  and  the  profile  of  the  flank  of  the  tooth  on  the  other  gear  is 
part  of  the  hypocycloid  traced  by  a  point  of  the  same  generating 
circle  rolling  on  the  inside  of  the  pitch  circle  of  the  second  gear, 
the  tooth  curves  fulfill  strictly  the  condition  required  for  tooth 
profiles  of  gear-wheels. 

Referring  again  to  Fig.  76,  the  generating  circle  GPE  rolling 
on  the  outside  of  the  pitch  circle  APB  can  only  generate  the  pro- 
file for  the  faces  of  the  teeth  on  the  lower  wheel,  and  rolling  on 
the  inside  of  CPD  it  can  only  generate  the  profile  for  the  flanks  of 
the  teeth  on  the  upper  wheel.  To  generate  the  profile  for  the 
flanks  of  the  teeth  on  the  lower  wheel  a  second  generating  circle 
MN  must  be  rolled  on  the  inside  of  the  pitch  circle  APB,  and  to 
generate  the  profile  for  the  faces  of  the  teeth  on  the  upper  wheel 
the  same  circle  MN  must  be  rolled  on  the  outside  of  CPD.  For 
two  wheels  to  gear  together  the  circular  pitch  must  be  the  same 
in  each  and  the  generating  circle  for  the  profile  of  the  faces  of  the 
teeth  on  the  one  must  be  the  same  size  as  the  generating  circle  for 
the  profile  of  the  flanks  of  the  teeth  on  the  other,  but  it  is  not 
necessary  that  the  same  generating  circle  be  used  for  the  profiles 


TOOTHED  GEARING  187 

of  the  faces  and  the  flanks  of  the  teeth  on  the  same  wheel.  It  is 
very  important,  however,  that  all  gear-wheels  of  the  same  circular 
pitch  shall  be  interchangeable,  and  to  accomplish  this  the  same 
generating  circle  must  be  used  for  the  profiles  of  the  faces  and 
flanks  of  the  teeth  on  all  the  wheels. 

If  the  diameter  of  the  generating  circle  is  larger  than  the  radius 
of  the  pitch  circle  the  profile  of  the  flank  of  the  tooth  will  curve 
rapidly  towards  the  center  of  the  tooth  at  the  root  circumference, 
thereby  diminishing  its  thickness  at  the  root  and  weakening  the 
tooth.  For  a  set  of  wheels  of  the  same  circular  pitch  required  to 
be  interchangeable  the  size  of  the  generating  circle,  which  must 
be  the  same  for  all,  is  limited  by  the  fact  that,  in  order  not  to  un- 
duly weaken  the  teeth  of  the  smallest  wheel  of  the  set,  the  diam- 
eter of  the  generating  circle  must  not  be  greater  than  the  pitch 
radius  of  that  wheel.  The  number  of  teeth  in  the  smallest  wheel 
of  the  cycloidal  system  is  taken  as  twelve,  and  for  all  wheels 
of  the  same  circular  pitch  the  diameter  of  the  generating 
circle  is  made  equal  to  the  pitch  radius  of  the  twelve-tooth 
wheel. 

Referring  to  Fig.  76,  suppose  the  upper  wheel  be  rotating 
counter-clockwise  in  the  direction  of  the  arrow.  It  will  then 
rotate  the  lower  wheel  clockwise  and,  if  G  be  the  first  point  of 
contact  between  the  flank  of  the  tooth  on  the  upper  wheel  and 
the  face  of  the  tooth  on  the  lower,  the  line  of  pressure  between  the 
teeth  at  that  instant,  neglecting  friction,  will  be  GP.  As  the 
wheels  advance  the  point  of  contact  of  the  teeth  will  slide  along 
the  arc  GP  of  the  generating  circle  GE,  and  the  direction  of  the 
line  of  pressure  will  continue  to  change  until  when  the  point  of 
contact  reaches  P  this  line  will  be  tangent  to  both  pitch  circles. 
After  the  point  P  has  been  reached  the  face  of  the  tooth  on  the 
upper  wheel  will  come  in  contact  with  the  flank  of  the  tooth  on 
the  lower,  and  the  point  of  contact  will  slide  along  the  arc  PJ  of 
the  generating  circle  MN,  the  direction  of  the  line  of  pressure 
between  the  teeth  changing  continually  until  the  teeth  are  about 
to  separate  at  the  point  J,  when  the  line  of  pressure  is  PJ.  The 
greatest  angle  between  the  common  tangent  to  the  pitch  circles 
and  the  line  of  pressure  between  the  teeth  occurs  when  contact  is 
just  commencing  or  ending.  This  angle  should  ordinarily  not  be 
permitted  to  be  greater  than  30°. 


188 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


As  in  the  case  of  the  involute  system,  the  arc  of  the  pitch  circle 
that  subtends  the  angle  through  which  a  wheel  rotates  from  the 
time  when  one  of  its  teeth  comes  in  contact  with  a  tooth  of  the 
other  wheel  until  the  point  of  contact  reaches  the  pitch  point  P 
is  called  the  arc  of  approach;  and  the  arc  of  the  pitch  circle  that 
subtends  the  angle  through  which  the  wheel  rotates  from  the  time 
the  point  of  contact  leaves  the  pitch  point  until  the  teeth  separate 
is  called  the  arc  of  recess.  The  sum  of  these  arcs  is  called  the  arc 
of  action.  In  order  that  there  shall  always  be  contact  between 
the  teeth  the  arc  of  action  should  be  at  least  equal  to  the  circular 
pitch.  If  it  is  desired  that  there  shall  always  be  contact  between 
two  pairs  of  teeth,  the  arc  of  action  should  be  at  least  equal  to 
twice  the  circular  pitch.  The  arc  of  action  may  be  increased  by 
increasing  the  length  of  the  teeth  up  to  a  certain  limit. 


Fig.  77. 

Fig.  77  shows  the  outline  of  the  cycloidal  tooth. 

94.  Spur  Gears.  —  The  gear-wheels  shown  in  Fig.  68  are  used 
to  transmit  rotary  motion  between  parallel  shafts.  The  pitch 
surfaces  of  the  wheels  are  cylindrical  and  the  tooth  surfaces  can 
be  generated  by  moving  the  involute  or  cycloidal  profile  in  a 
direction  parallel  to  the  axis  of  the  shaft.  Gear-wheels  used  to 
transmit  rotary  motion  between  parallel  shafts  are  called  spur 
gears.  The  wheel  that  imparts  motion  to  the  other  is  called  the 
driver  and  the  driven  wheel  is  called  the  follower.  Either  one  of 
a  pair  of  spur  gears  in  mesh  may  be  the  driver. 


TOOTHED  GEARING 


189 


GO 
•O 


190 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


Rack  and  Spur  Gear.  —  If  the  radius  of  one  of  the  spur  gears 
shown  in  Fig.  68  be  increased  until  it  becomes  infinite,  the  spur 
gear  becomes  a  rack  and  its  pitch  circumference  a  straight  line 
whose  linear  velocity  is  the  same  as  that  of  a  point  on  the  pitch 
circumference  of  the  gear  with  which  it  meshes.  A  rack  and  spur 
gear  are  generally  used  to  change  a  reciprocating  rotary  motion 
into  a  reciprocating  motion  of  translation  or  vice  versa. 

Either  the  involute  or  the  cycloidal  system  may  be  used  for 
the  teeth  of  a  rack  and  spur  gear.  In  the  involute  system  the 
profile  of  the  rack  tooth  is  a  straight  line  perpendicular  to  the  line 
of  action  to  which  the  base  circle  of  the  spur  gear  is  tangent.  In 
the  cycloidal  system  the  profile  of  the  face  of  the  rack  tooth  is  a 
cycloid  traced  by  a  point  on  the  generating  circle  while  rolling  on 
the  outside  of  the  pitch  line  of  the  rack.  The  profile  of  the  flank 
of  the  rack  tooth  is  also  a  cycloid  traced  by  a  point  of  the  generat- 
ing circle  while  rolling  on  the  inside  of  the  pitch  line  of  the  rack. 

For  interchangeability  the  diameter  of 
the  generating  circle  is  made  equal  to 
the  pitch  radius  of  the  twelve-tooth 
gear  of  the  same  circular  pitch.  Either 
the  spur  gear  or  the  rack  may  be  the 
driver. 

Fig.  78  shows  a  rack  and  spur  gear. 
95.  Bevel  Gears.  —  If  the  shafts 
to  be  connected  by  gear-wheels  would 
intersect  if  prolonged,  the  pitch  sur- 
faces cannot  be  cylindrical  but  must 
be  the  surfaces  of  frustums  of  cones 
whose  apexes  are  at  the  point  of  inter- 
section of  the  shafts,  as  shown  in 
Fig.  79,  in  which  AB  and  CD  are  the 
shafts  which  if  prolonged  would  inter- 
sect at  E,  and  FGHI  and  HIJK  are 
the  frustums  of  cones  whose  apexes  are 
at  E  and  whose  surfaces  are  the  pitch 
surfaces  of  gear-wheels  transmitting  rotary  motion  between  the 
shafts.  Gear-wheels  with  conical  pitch  surfaces  transmitting 
rotary  motion  between  shafts  that  would  intersect  if  prolonged 
are  called  bevel  gears.  The  angular  velocities  of  a  pair  of  bevel 


Fig.  79. 


TOOTHED  GEARING  191 

gears  in  mesh  are  inversely  proportional  to  the  sines  of  the  angles 
between  the  elements  of  the  pitch  surfaces  in  contact  and  the 
shafts  to  which  the  gears  are  attached  or,  more  simply,  inversely 
proportional  to  the  numbers  of  the  teeth  on  the  gears.  Either 
one  of  a  pair  of  bevel  gears  in  mesh  may  be  the  driver. 

Both  the  involute  and  the  cycloidal  systems  are  used  for  the 
teeth  of  bevel  gears  but  the  tooth  surfaces,  while  having  rectilinear 
elements,  cannot  be  formed  by  sliding  a  profile  of  the  tooth  along 
a  straight  line.  For  the  involute  system  the  tooth  surfaces  are 
obtained  by  rolling  a  plane  on  a  base  cone  whose  apex  is  at  the 
point  of  intersection  of  the  shafts,  and  for  the  cycloidal  system 
the  tooth  surfaces  are  obtained  by  rolling  generating  cones  on  the 
inside  and  outside  of  the  pitch  cones.  The  tooth  profile  of  a  bevel 
gear  at  any  section  at  right  angles  to  the  axis  of  the  shaft  can  of 
course  be  obtained  by  rolling  a  line  on  the  base  circle  cut  by  the 
section  from  the  base  cone,  or  by  rolling  a  generating  circle  on 
the  inside  and  outside  of  the  pitch  circle  cut  by  the  section  from 
the  pitch  cone. 

Fig.  80  shows  a  pair  of  bevel  gears  in  mesh. 

96.  Screw  Gearing.  —  Worm  and  Worm- Wheel.  —  Referring 
to  Fig.  78,  showing  a  rack  and  spur  gear,  suppose  the  teeth  of  the 
rack  instead  of  being  parallel  to  the  axis  of  the  gear  be  given  an 
inclination  to  that  axis  as  shown  in  Fig.  81.  In  order  that  the 
teeth  of  the  gear  shall  mesh  properly  with  the  inclined  teeth  of 
the  rack  the  former  must  be  inclined  also  as  shown  in  elevation 
in  Fig.  81,  what  were  formerly  rectilinear  elements  of  the  teeth 
now  becoming  helices  on  the  cylindrical  surface  of  the  gear.  The 
inclination  of  the  teeth  will  in  no  way  affect  the  working  of  the 
rack  and  gear  and  a  longitudinal  movement  of  the  rack  at  right 
angles  to  the  axis  of  the  gear  will  cause  rotation  of  the  latter  as 
before.  Owing  to  the  inclination  of  the  teeth,  however,  a  move- 
ment of  the  rack  at  right  angles  to  the  axis  of  the  gear  is  not  the 
only  movement  of  the  rack  that  will  rotate  the  gear,  for  an  ex- 
amination of  Fig.  81  shows  at  once  that  a  movement  of  the  rack 
in  a  direction  parallel  to  the  axis  of  the  gear,  that  is,  perpendicu- 
lar to  the  plane  of  the  paper,  will  also  cause  rotation  of  the  gear 
which  will,  however,  be  rather  limited  because  of  the  compara- 
tively limited  width  of  the  rack. 

In  the  figure  the  distance  ab  is  the  pitch  of  the  rack  teeth, 


192 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


TOOTHED  GEARING 


193 


equal  to  the  circular  pitch  of  the  teeth  of  the  gear,  and  ac  is  equal 
to  the  length  of  the  arc  measured  on  its  pitch  circumference 
through  which  the  gear  would  be  rotated  by  a  movement  of  the 
rack  over  a  distance  equal  to  its  width  in  a  direction  parallel  to 


Plan  of  Rack. 


Toothr/     ^Space. 


Fig.  81. 

the  axis  of  the  gear.  Now  suppose  the  rack  be  rolled  into  a 
cylinder  whose  axis  is  parallel  to  the  longest  dimension  of  the 
original  rack  and  whose  circumference  is  equal  to  the  width  of  the 
rack;  then  a  tooth  on  the  rack  will  become  one  complete  turn  of  a 
helical  projection  or  tooth  on  the  cylinder,  and  the  pitch  of  the 
helix  so  formed  will  be  equal  to  the  distance  ac.  If  all  teeth  but 
one  be  removed  from  the  cylinder  and  the  latter  be  placed  with  its 
axis  perpendicular  to  that  of  the  gear  and  so  that  one  end  of  the 
helix  formed  by  the  remaining  tooth  engages  in  a  tooth  space  of 
the  gear,  the  cylinder  can  be  rotated  about  its  axis  and  its  helical 
tooth  during  one  complete  revolution  will  cause  the  gear  to  rotate 
through  an  arc  measured  on  its  pitch  circumference  equal  to  the 
distance  ac,  Fig.  81,  in  the  same  manner  as  did  the  inclined  tooth 


194  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

on  the  rack  when  the  latter  was  moved  over  a  distance  equal  to 
its  width  parallel  to  the  axis  of  the  gear.  If  the  rack  is  of  the 
width  shown  in  the  figure,  it  is  evident  that  after  one  revolution 
of  the  cylinder  formed  therefrom  further  motion  will  be  prevented 
by  the  end  of  its  helical  tooth  coming  against  the  end  of  one  of  the 
teeth  on  the  gear,  for  the  movement  of  the  latter  measured  on  its 
pitch  circumference  has  been  less  than  its  circular  pitch.  If, 
however,  the  width  of  the  rack  be  increased  so  that  the  distance 
ac,  Fig.  81,  which  is  the  amount  one  end  of  an  inclined  tooth  is  in 
advance  of  the  other  end,  is  just  equal  to  ab,  the  circular  pitch  of 
the  gear,  and  the  new  rack  be  rolled  into  a  cylinder,  it  will  be  seen 
that,  when  the  helical  tooth  on  the  cylinder  is  engaged  in  a  tooth 
space  of  the  gear  as  before  and  the  cylinder  is  rotated  once  about 
its  axis,  the  gear  will  be  rotated  through  an  arc  measured  on  its 
pitch  circumference  equal  to  its  circular  pitch.  After  one  com- 
plete revolution  of  the  cylinder  the  end  of  its  helical  tooth  will 
occupy  exactly  the  same  position  as  when  it  was  first  engaged 
with  a  tooth  space  on  the  gear,  but  that  space  will  have  moved 
through  an  arc  equal  to  the  circular  pitch  of  the  gear  so  that  the 
helical  tooth  can  no  longer  engage  with  it;  but  as  the  distance 
between  the  centers  of  the  tooth  spaces  of  the  gear  is  equal  to  its 
circular  pitch,  the  space  in  rear  will  occupy  exactly  the  same  posi- 
tion as  did  the  first  before  the  cylinder  was  rotated  and,  conse- 
quently, the  end  of  the  helical  tooth  on  the  cylinder  will  now 
engage  with  it.  Another  complete  revolution  of  the  cylinder 
will  again  rotate  the  gear  through  an  arc  measured  on  its  pitch 
circumference  equal  to  its  circular  pitch,  will  restore  the  helical 
tooth  on  the  cylinder  to  its  original  position,  and  will  bring  a  third 
tooth  space  of  the  gear  in  position  to  be  engaged  by  the  tooth  of 
the  cylinder,  and  so  on;  so  that  continuous  rotation  of  the  cylinder 
will  cause  continuous  rotation  of  the  gear. 

In  this  combination  the  gear  with  helical  teeth  is  called  a  worm- 
wheel  and  the  cylindrical  rack  with  a  helical  tooth  is  called  a 
worm.  A  worm  and  worm-wheel  form  a  particular  case  of  screw 
gearing,  so  called  because  of  the  screw-like  action  of  the  teeth  of 
the  gears. 

In  this  diseussion  reference  has  been  had  to  a  length  of  the 
helical  tooth  on  the  worm  sufficient  to  make  one  complete  turn 
only  about  its  axis.  This  length  is  all  that  is  necessary  to  secure 


TOOTHED  GEARING 


195 


Fig.  82.  —  Worm  and  Worm-Wheel. 


196 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


continuous  rotation  of  the  worm-wheel,  but  in  practice  several 
turns  are  used  as  this  distributes  the  pressure  over  two  or  more 
teeth  on  the  wheel. 

Fig.  82  shows  a  worm  and  worm-wheel  in  mesh. 

Referring  again  to  Fig.  81,  it  is  evident  that  the  rack  with 
inclined  teeth  need  not  be  perpendicular  to  the  axis  of  the  spur 
gear  provided  the  helical  teeth  on  the  gear  make  the  same  angle 
with  its  axis  as  do  the  inclined  teeth  on  the  rack.  Such  a  rack 
and  spur  gear  are  shown  in  plan  in  Fig.  83,  the  teeth  of  the  rack 
being  marked  A  and  those  of  the  gear  being  marked  B. 


NOTE.  —  The  outlines  of  the  teeth  shown  in  this  figure  correspond  to  sections 
of  the  teeth  at  the  pitch  surfaces. 

Fig.  83. 

When  this  combination  is  used  as  a  rack  and  gear  the  rack 
must  be  held  in  guideways  to  compel  it  to  move  only  in  the  direc- 
tion of  its  longest  dimension,  and  the  gear  must  be  prevented 
from  moving  in  a  direction  parallel  to  the  shaft  by  collars  on  the 
shaft  or  other  suitable  means.  If  the  guideways  be  removed, 


TOOTHED  GEARING  197 

however,  and  the  rack  given  a  motion  of  translation  at  right  angles 
to  the  direction  of  its  longest  dimension  it  will  still  cause  a  limited 
rotation  of  the  gear;  and,  by  giving  the  rack  the  proper  width 
and  wrapping  it  around  a  cylinder  whose  axis  is  parallel  to  the 
longest  dimension  of  the  rack,  it  will  become  a  worm  as  before 
which,  when  rotated  continuously  about  its  axis,  will  cause  con- 
tinuous rotation  of  the  gear,  now  called  a  worm-wheel.  It  is 
evident  from  this  that  it  is  not  necessary  that  the  axes  of  a  worm 
and  worm-wheel  be  at  right  angles  to  each  other,  although  such 
is  generally  the  case. 

In  Fig.  83  the  circular  pitch  of  the  gear  is  equal  to  the  distance 
ac,  which  is  the  distance  between  corresponding  points  of  adjacent 
teeth  of  the  rack  measured  in  a  direction  at  right  angles  to  the 
axis  of  the  gear.  Now  in  order  to  rotate  the  gear  through  an  arc 
measured  on  its  pitch  circumference  equal  in  length  to  ac  by  a 
movement  of  the  rack  at  right  angles  to  its  longest  dimension  over 
a  distance  equal  to  its  width,  it  is  evident  that  the  width  of  the 
rack  must  be  made  equal  to  the  distance  cd,  the  point  d  being  the 
intersection  of  the  edge  of  the  tooth  passing  through  a  by  a  line 
drawn  through  c  at  right  angles  to  the  longest  dimension  of  the 
rack.  It  is  also  apparent  that,  when  the  width  of  the  rack  is 
made  equal  to  cd  and  the  rack  is  rolled  into  a  cylinder  whose 
axis  is  ef,  the  various  teeth  of  the  rack  will  form  so  many  com- 
plete turns  of  a  continuous  helical  tooth  on  the  cylinder,  and  that 
one  complete  revolution  of  the  cylinder  or  worm  will  rotate  the 
gear,  now  called  a  worm-wheel,  through  an  arc  measured  on  its 
pitch  circumference  equal  to  its  circular  pitch,  which  is  equal  to 
ac.  The  pitch  of  the  helix  on  the  worm  will  then  be  equal  to  be, 
and  the  circular  pitch  of  the  wheel,  equal  to  ac,  is  equal  to  the 
oblique  projection  on  a  plane  at  right  angles  to  the  axis  of  the 
wheel  of  the  pitch  of  the  helix  of  the  worm,  the  projecting  lines 
being  parallel  to  the  teeth  of  the  wheel. 

Let  eg  be  a  line,  in  a  plane  parallel  to  the  axes  of  the  worm  and 
wheel,  drawn  perpendicular  to  the  teeth  through  their  point  of 
contact  c,  and  let  a  be  the  angle  which  this  line  makes  with  the 
axis  ef  of  the  worm,  and  0  be  the  angle  which  it  makes  with  a  plane 
at  right  angles  to  the  axis  of  the  wheel;  then  if  a  is  greater  than  0 
a  point  on  the  pitch  circumference  of  the  wheel  will,  during  one 
complete  revolution  of  the  worm,  move  through  an  arc  less  in 


198  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

length  than  the  pitch  of  the  helix  on  the  worm,  and  vice  versa  if  a 
is  less  than  /3.  If  a  is  equal  to  /3  a  point  on  the  pitch  circumference 
of  the  wheel  will,  during  one  complete  revolution  of  the  worm, 
move  through  an  arc  equal  in  length  to  the  pitch  of  the  helix  on 
the  worm,  the  same  as  it  would  if  the  axes  of  the  worm  and  wheel 
were  perpendicular  to  each  other. 

Since  the  axis  of  the  worm  is  oblique  to  the  axis  of  the  wheel, 
the  point  of  contact  of  the  teeth  during  one  revolution  of  the  worm 
travels  in  a  diagonal  line  across  the  rim  of  the  wheel  instead  of 
remaining  in  its  central  plane  as  it  does  when  the  axes  of  the  worm 
and  wheel  are  perpendicular  to  each  other.  On  this  account  the 
rim  of  the  wheel  must  not  be  too  narrow  or  the  teeth  will  not 
remain  in  contact  during  one  complete  revolution  of  the  worm. 

Spiral  Gears.  —  If  instead  of  wrapping  the  rack  shown  in 
Fig.  83  around  a  cylinder  whose  axis  is  parallel  to  the  longest 
dimension  of  the  rack,  it  be  wrapped  around  one  whose  axis  is 
perpendicular  to  that  dimension,  the  continuous  rotation  of  this 
cylinder  will  also  cause  continuous  rotation  of  the  gear,  and  the 
combination  so  formed  is  called  a  pair  of  spiral  gears.  Fig.  84 
shows  a  pair  of  spiral  gears  in  mesh. 

Spiral  gears  are  also  a  particular  form  of  screw  gearing  and  do 
not  differ  in  principle  from  a  worm  and  worm-wheel  as  may  be 
inferred  from  the  manner  in  which  they  are  developed  from  a  rack 
and  spur  gear,  or  as  may  perhaps  be  more  readily  seen  from  the 
following  discussion.  Suppose  the  cylinder  formed  by  wrapping 
the  rack  of  Fig.  83  around  an  axis  perpendicular  to  its  longest 
dimension  be  increased  in  length  until  one  of  its  helical  teeth  makes 
a  complete  turn  about  it,  and  that  all  teeth  except  this  one  be 
removed  from  the  cylinder.  Then,  except  for  the  fact  that  the 
pitch  of  the  helix  on  the  cylinder  is  relatively  so  great  in  compari- 
son with  the  diameter  and  width  of  the  gear  that  the  tooth  of 
the  gear  would  probably  rotate  out  of  contact  with  that  of  the 
cylinder  before  the  latter  could  make  one  complete  revolution, 
one  such  revolution  of  the  cylinder  would  cause  a  point  on  the 
pitch  circumference  of  the  gear  to  move  through  an  arc  equal  in 
length  to  the  oblique  projection  on  a  plane  at  right  angles  to  the 
axis  of  the  gear  of  the  pitch  of  the  helix  on  the  cylinder,  the  pro- 
jecting lines  being  parallel  to  the  teeth  of  the  gear;  and,  further- 
more, by  increasing  the  diameter  of  the  gear,  keeping  the  inclina- 


TOOTHED  GEARING 


199 


Fig.  84.  —  Spiral  Gears. 


200  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

tion  of  its  helical  teeth  the  same,  the  curvature  of  its  pitch  circle 
can  be  reduced  to  such  an  extent  that,  when  the  width  of  the  gear 
is  sufficiently  increased,  its  tooth  will  remain  in  contact  with  that 
of  the  cylinder  during  one  complete  revolution  of  the  latter. 

It  is  thus  seen  that  by  increasing  the  length  of  one  of  the 
spiral  gears  developed  from  the  rack  and  spur  gear  of  Fig.  83 
and  increasing  the  diameter  and  the  width  of  the  other,  keeping 
the  inclination  of  the  helical  teeth  the  same  in  both  cases,  a  com- 
bination has  been  produced  in  which,  if  the  first  gear  be  rotated 
through  a  complete  revolution,  the  other  will  be  rotated  through 
an  arc  measured  on  its  pitch  circumference  equal  in  length  to 
the  oblique  projection  on  a  plane  at  right  angles  to  its  axis  of 
the  pitch  of  the  helix  of  the  first;  and,  if  the  circular  pitch  of  the 
teeth  on  the  second  gear  be  made  equal  to  this  arc,  continuous 
rotation  of  the  first  gear  will  cause  continuous  rotation  of  the 
second.  This  combination  has  now  become  a  worm  and  worm- 
wheel  differing  in  no  essential  from  the  worm  and  wheel  formed 
by  wrapping  the  rack  of  Fig.  83  around  a  cylinder  whose  axis  is 
parallel  to  the  longest  dimension  of  the  rack;  and  this  transfor- 
mation has  been  effected  without  changing  the  angle  between  the 
axes  of  the  gears  or  those  between  the  teeth  and  the  axes,  but 
only  by  increasing  the  length  of  one  gear  and  the  diameter,  width, 
and  circular  pitch  of  the  other,  which  changes  evidently  do  not 
affect  the  principle  upon  which  the  gears  operate. 

97.  Distinction  between  Worm- Gearing  and  Spiral  Gearing.  - 
If  the  helical  teeth  formed  by  wrapping  the  rack  around  a  cylinder 
make  so  great  an  angle  with  its  axis  that  they  make  one  or  more 
complete  turns  around  it  in  the  length  of  the  cylinder,  the  result- 
ing gear  is  called  a  worm  and  the  gear-wheel  with  which  it  meshes 
is  called  a  worm-wheel.  By  examination  of  Figs.  81  and  83  it  is 
seen  that  the  helical  teeth  formed  by  wrapping  the  rack  around 
a  cylinder  whose  axis  is  parallel  to  the  longest  dimension  of  the 
rack  make  several  turns  around  that  axis  in  the  length  of  the 
cylinder,  and  this  is  why  the  resulting  gear  was  called  a  worm 
and  that  with  which  it  meshes,  a  worm  wheel. 

If  the  helical  teeth  formed  by  wrapping  the  rack  around  a 
cylinder  make  so  small  an  angle  with  its  axis  that  they  do  not 
make  a  complete  turn  about  it  in  the  length  of  the  cylinder,  the 
resulting  gear  and  the  one  with  which  it  meshes  are  called  spiral 


ENGINEERING  BURh 
SJ3GTI9S 


TOOTHED  GEARING  201 

gears.  By  examination  of  Fig.  83  it  is  seen  that  when  the  rack  is 
wrapped  around  a  cylinder  whose  axis  is  perpendicular  to  the 
longest  dimension  of  the  rack,  the  helical  teeth  so  formed  will  make 
but  a  small  portion  of  a  turn  around  the  axis  of  the  cylinder.  For 
this  reason  the  resulting  gear  and  that  with  which  it  meshes  were 
called  spiral  gears. 

98.  Velocity  Ratio  of  Worm-  Gears.  —  It  has  already  been 
shown  that  by  choosing  a  suitable  width  for  the  rack  shown  in 
Fig.  81  and  then  rolling  it  into  a  cylinder  whose  axis  is  parallel 
to  the  longest  dimension  of  the  rack,  the  pitch  of  the  helix  on  the 
cylinder  may  be  made  equal  to  the  circular  pitch  of  the  gear. 
Similarly,  by  choosing  a  suitable  width  for  the  rack  shown  in 
Fig.  83  and  rolling  it  into  a  cylinder  whose  axis  is  parallel  to  the 
longest  dimension  of  the  rack,  the  pitch  of  the  helix  on  that 
cylinder  may  be  made  of  such  a  length  that  its  oblique  projection 
on  a  plane  at  right  angles  to  the  axis  of  the  gear,  by  lines  parallel 
to  the  teeth  of  the  gear,  will  be  equal  to  the  circular  pitch  of  the 
gear.  Each  separate  tooth  of  either  rack  will  then  form  one  com- 
plete turn  of  a  helix  on  the  cylinder  and,  furthermore,  the  various 
turns  formed  by  the  teeth  of  the  rack  will  become  so  many  parts 
of  one  continuous  helical  tooth  or  thread  on  the  cylinder,  so  that 
the  resulting  worm  is  called  a  single-threaded  worm. 

One  revolution  of  a  single-threaded  worm  will  rotate  the 
worm-wheel  through  an  arc  measured  on  its  pitch  circumference 
equal  in  length  to  its  circular  pitch,  and  to  cause  the  wheel  to 
make  a  complete  revolution  the  worm  must  make  as  many  revo- 
lutions as  there  are  teeth  on  the  wheel.  Since  the  single  thread 
on  the  worm  is  in  reality  a  single  tooth,  the  angular  velocities  of 
the  worm  and  worm-wheel  are  to  each  other  inversely  as  the 
numbers  of  their  teeth,  as  is  the  case  with  all  toothed  gears. 

If  the  width  of  the  rack  be  increased  to  twice  its  former  amount, 
the  inclination  of  the  teeth  remaining  the  same  as  before,  every 
alternate  tooth  of  the  rack,  when  it  is  rolled  into  a  cylinder 
whose  axis  is  parallel  to  that  of  the  first  cylinder,  will  form  part 
of  a  continuous  helical  tooth  or  thread  around  the  cylinder;  and 
there  will  be  two  such  threads,  the  worm  in  this  case  being  called 
a  double-threaded  worm.  Since  the  pitch  of  the  helix  on  the  worm 
is  double  what  it  was  before,  either  one  of  the  two  threads  will 
engage  only  with  every  alternate  tooth  of  the  gear,  the  remaining 


202  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

teeth  of  the  gear  engaging  with  the  other  thread  on  the  worm. 
One  revolution  of  a  double-threaded  worm  will  rotate  the  worm- 
wheel  through  an  arc  measured  on  its  pitch  circumference  equal 
in  length  to  twice  its  circular  pitch,  and  to  cause  the  wheel  to 
make  a  complete  revolution  the  worm  must  make  half  as  many 
revolutions  as  there  are  teeth  on  the  wheel. 

If  there  are  three  separate  threads  on  the  worm  it  is  called  a 
triple-threaded  worm,  and  so  on.  When  any  multiple-threaded 
worm  meshes  with  a  worm-wheel  the  angular  velocities  of  the 
worm  and  wheel  are  inversely  as  the  numbers  of  the  teeth  or 
threads  on  the  worm  and  the  worm-wheel.  The  ratio  of  the 
angular  velocities  of  a  worm  and  worm-wheel  in  mesh  is  inde- 
pendent of  their  radii.  The  teeth  on  a  worm  are  referred  to  as 
threads. 

Velocity  Ratio  of  Spiral  Gears.  —  The  angular  velocities  of 
a  pair  of  spiral  gears  in  mesh  are  inversely  as  the  numbers  of  the 
teeth  on  the  gears,  and  the  ratio  of  the  angular  velocities  is  in- 
dependent of  the  radii  of  the  gears  except  in  the  one  case  where 
the  axes  of  the  gears  are  perpendicular  to  each  other  and  the 
helix  angle  between  the  teeth  of  each  gear  and  a  line  perpen- 
dicular to  its  axis  is  forty-five  degrees.  In  this  case  the  numbers 
of  the  teeth  on  the  gears  are  directly  proportional  to,  and  the 
angular  velocities  inversely  proportional  to,  the  pitch  radii  of  the 
gears. 

Tooth  Curves  of  Screw  Gears.  —  In  view  of  the  development 
of  screw  gearing  from  a  rack  and  spur  gear  it  is  evident  that  the 
tooth  surfaces  may  belong  to  either  the  involute  or  the  cycloidal 
system.  The  involute  system  is  generally  preferred  for  worms 
because  the  involute  thread  for  racks  and  worms  has  straight 
sides,  and  is  on  this  account  more  readily  cut  in  the  lathe  than 
the  cycloidal  thread. 

99.  Shafts  Connected  by  Screw  Gearing.  —  Drivers.  —  Pitch 
of  Screw  Gearing.  —  Since  the  axes  of  spiral  gears  and  worm 
and  worm-wheels  in  mesh  are  in  different  planes  and  not  parallel 
to  each  other,  such  gearing  is  used  to  transmit  rotary  motion 
between  shafts  that  are  neither  parallel  nor  intersecting.  The 
worm  and  worm-wheel  are  used  particularly  when  it  is  desired 
to  obtain  a  large  velocity  ratio  between  the  shafts. 

Either  one  of  a  pair  of  spiral  gears  in  mesh  may  be  the  driver. 


TOOTHED  GEARING  203 

In  practice,  however,  that  spiral  gear  is  taken  as  the  driver  which 
has  the  smaller  helix  angle  between  its  teeth  and  a  line  perpen- 
dicular to  its  axis,  since  greater  efficiency  of  the  gears  results  from 
doing  so. 

In  the  case  of  a  worm  and  worm-wheel  the  worm  is  the  driver. 
When  the  angle  between  the  threads  of  the  worm  and  a  line 
perpendicular  to  its  axis  is  less  than  the  angle  of  friction,  which 
is  generally  the  case,  the  worm-wheel  cannot  drive  the  worm. 
If,  however,  this  angle  is  greater  than  the  angle  of  friction  the 
worm-wheel  can  drive  the  worm. 

The  circular  pitch  of  a  spiral  gear  is  the  distance  between  cor- 
responding points  of  adjacent  teeth  measured  on  the  cylindrical 
pitch  surface  at  right  angles  to  the  axis  of  the  gear.  The  distance 
measured  on  the  pitch  surface  at  right  angles  to  the  teeth  is  the 
normal  pitch,  and  that  measured  in  a  direction  parallel  to  the 
axis  of  the  gear  is  the  axial  pitch.  For  two  spiral  gears  to  run 
together  the  normal  pitches  must  be  the  same  in  both,  and,  if  the 
axes  of  the  gears  are  perpendicular  to  each  other,  the  circular 
pitch  of  the  one  must  also  be  equal  to  the  axial  pitch  of  the  other. 

The  term  circular  pitch  is  not  used  in  connection  with  a  worm, 
but  it  is  in  connection  with  a  worm-wheel  and  has  the  same 
significance  with  regard  to  the  latter  as  for  a  spiral  gear.  The 
axial  pitch  of  a  worm  is  the  distance  measured  on  its  pitch  surface 
in  a  direction  parallel  to  its  axis  between  corresponding  points 
of  adjacent  threads  of  a  multiple-threaded  worm  or  of  adjacent 
parts  of  the  same  thread  of  a  single-threaded  worm.  In  the  case 
of  a  single-threaded  worm  the  axial  pitch  is  the  same  as  the 
pitch  of  the  helix  formed  by  its  thread.  The  distance  between 
corresponding  points  of  adjacent  threads  of  a  multiple-threaded 
worm  or  of  adjacent  parts  of  the  same  thread  of  a  single-threaded 
worm  measured  on  its  pitch  surface  at  right  angles  to  the  threads 
is  the  normal  pitch.  The  axial  and  normal  pitches  of  a  worm- 
wheel  are  the  same  as  for  a  spiral  gear.  For  a  worm  and  worm- 
wheel  to  run  together  the  normal  pitches  must  be  the  same  in 
both  and,  when  their  axes  are  perpendicular  to  each  other,  as  is 
generally  the  case,  the  axial  pitch  of  the  worm  must  be  equal  to 
the  circular  pitch  of  the  worm-wheel. 

The  lead  of  a  worm  is  the  distance  any  separate  thread  ad- 
vances in  the  direction  of  the  axis  of  the  worm  while  making  one 


204 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


complete  turn  around  it.    It  is  the  same  as  the  pitch  of  the  helix 
formed  by  the  thread. 

100.  Wheel  Train.  —  Velocity  Ratio  of  First  and  Last  Shafts 
Connected  by  a  Wheel  Train.  —  A  series  of  gears  interposed 
between  two  shafts  is  called  a  wheel  or  gear  train.  In  Pig.  85 


Fig.  85. 

is  shown  a  wheel  train  connecting  the  driving  shaft  a  with  the 
driven  shaft  j. 

A  spur  gear  A  on  the  shaft  a  meshes  with  another  B.  On  the 
same  shaft  with  B  is  a  bevel  gear  C  meshing  with  another  bevel 
gear  D  carried  by  a  third  shaft,  on  the  other  end  of  which  is  a 


TOOTHED  GEARING  205 

spiral  gear  E.  The  gear  E  meshes  with  a  second  spiral  gear  F 
and  on  the  same  shaft  with  F  is  a  single-threaded  worm  G 
driving  a  worm-wheel  H.  On  the  other  end  of  the  shaft  which 
carries  H  is  a  spur  gear  /  that  drives  the  spur  gear  J  on  the 
shaft  j. 

Let  it  be  required  to  find  the  ratio  of  the  angular  velocities  of 
the  shafts  a  and  j  or,  what  is  the  same  thing,  the  ratio  of  the 
numbers  of  revolutions  per  minute  of  the  shafts.  Let  Na  repre- 
sent the  number  of  revolutions  per  minute  of  the  shaft  a  and  of 
the  gear  A,  Nb  the  number  of  revolutions  per  minute  of  the  gear 
B,  Nc  the  number  of  revolutions  per  minute  of  the  gear  C,  etc.; 
and  let  Ta  represent  the  number  of  teeth  of  A,  Tb  the  number  of 
teeth  of  B,  etc.  Then 

#•_?*.  ^  _?V  **i-Tf.  **i-Tk    N_i_Ti 
Nb     Ta'  Nd  ~  Tc'  Nf  ~  Te'  Nh  ~  Tg'  N,-  ~  Tt 

Multiplying  together  the  first  terms  of  the  various  equalities 
and  placing  the  product  equal  to  the  product  of  all  the  last  terms 
of  the  equalities,  we  have 

NaxNcxNexN0X  Ni      TbxTdxTfxThx  T; 


NbxNdXNfxNhX  N,      Ta  X  Tcx  TexTgx  Tt 

But  since  B  and  C  are  on  the  same  shaft  Nb  =  Nc,  and,  similarly, 
Nd  =  Ne,  Nf  =  Na,  and  Nh  =  Ni.  Making  these  substitutions  in 
equation  (1)  we  obtain 

Na  _  TbxTdxTfxThx  T, 
N}-     TaxTcxTexT0x  Ti 

Referring  to  Fig.  85  it  will  be  seen  that  the  gears  B,  D,  F,  H, 
and  J  are  all  driven  gears  or  followers  and  the  gears  A,  C,  E,  G, 
and  /  are  all  drivers,  and  we  may,  therefore,  write  that  when  two 
shafts  are  connected  by  a  wheel  train  the  ratio  of  the  angular  velocities 
of  the  driving  and  driven  shafts  or  of  the  numbers  of  revolutions  per 
minute  of  these  shafts  is  equal  to  the  product  of  the  numbers  of  the 
teeth  of  all  the  followers  divided  by  the  product  of  the  numbers  of  the 
teeth  of  all  the  drivers. 

101.  Idlers.  —  If  a  wheel  train  connecting  two  shafts  consists 
of  three  or  more  gears  meshing  directly  the  one  into  the  other  as 


206 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


shown  in  Fig.  86,  the  velocity  ratio  of  the  driving  and  driven 
shafts  is  the  same  as  if  the  gears  on  these  shafts  meshed  directly 
into  each  other  without  the  interposition  of  the  intermediate 
gears,  for  the  gear  B  is  a  follower  with  respect  to  A  and  a  driver 
with  respect  to  C  and  the  number  of  its  teeth  will  appear  both  in 
the  numerator  and  the  denominator  of  the  fraction  representing 


A 


a 


B 


Fig.  86. 

the  ratio  of  the  angular  velocities  of  the  shafts  a  and  c,  and  will, 
consequently,  cancel  out  of  the  fraction.  If  another  gear  be  in- 
terposed between  B  and  C  it  also  will  act  both  as  a  follower  and 
a  driver  and  can,  consequently,  have  no  effect  on  the  velocity 
ratio  of  a  and  c. 

By  examination  of  Fig.  86  it  will  be  seen  that  gears  A  and  B 
turn  in  opposite  directions  as  do  gears  B  and  C  and,  consequently, 
A  and  C  and  the  shafts  to  which  they  are  fastened  turn  in  the 
same  direction.  If  another  gear  be  interposed  between  A  and  C 
the  latter  will  turn  in  the  opposite  direction  from  A;  and,  in 
general,  if  the  number  of  gears  meshing  directly  the  one  into  the 
other  as  shown  in  Fig.  86  is  odd,  the  first  and  last  gears  will  turn 


TOOTHED  GEARING  207 

in  the  same  direction,  but  if  the  number  is  even  the  first  and  last 
gears  will  turn  in  opposite  directions.  All  the  gears  in  such  a 
train  except  the  first  and  last  are  called  idlers,  and  their  function 
is  principally  to  change  the  direction  of  rotation  of  the  driven 
shaft  since  they  cannot  change  its  angular  velocity.  In  an  ex- 
ceptional case  when  shafts  are  so  far  apart  that  two  gears  con- 
necting them  would  be  inconveniently  large,  idlers  might  be  used 
to  bridge  over  the  distance  between  them. 

102.  Relation  between  the  Power  and  the  Resistance  in  a 
Wheel  Train.  —  Referring  to  Fig.  85,  the  shaft  a  is  driven  by 
some  source  of  energy  which  applies  a  force  P  called  the  power 
at  the  end  of  a  lever  arm  p  with  respect  to  the  axis  of  the  shaft. 
The  source  of  energy  may  be  human  as  when  the  shaft  a  is  turned 
by  the  hand  at  the  extremity  of  the  lever  p,  or  the  shaft  may  be 
driven  by  still  another  gear-wheel  not  shown  in  the  figure,  or  by 
a  belt.  If  driven  by  another  gear-wheel  the  power  will  be  the 
force  exerted  between  the  teeth  of  the  wheels  and  its  lever  arm 
may  be  taken  as  the  pitch  radius  of  the  driven  wheel  on  the  shaft 
a.  If  the  shaft  is  driven  by  a  belt  the  power  will  be  the  tension 
in  the  belt,  or,  more  strictly,  the  difference  in  tension  between  the 
upper  and  lower  parts  of  the  belt  considered  as  horizontal,  and 
its  lever  arm  will  be  the  radius  of  the  pulley  increased  by  one- 
half  the  thickness  of  the  belt.  The  energy  received  from  the 
source  is  transmitted  through  the  wheel  train  and,  neglecting 
friction,  is  all  delivered  to  the  shaft  j  where  it  is  made  to  perform 
useful  work  in  overcoming  a  resistance  R  having  a  lever  arm  r 
with  respect  to  the  axis  of  the  shaft  j. 

Let  it  be  required  to  find  the  power  P  that  must  be  applied 
with  a  lever  arm  p  to  the  shaft  a  to  overcome  through  the  wheel 
train  shown  in  Fig.  85  a  resistance  R  having  a  lever  arm  r  with 
respect  to  the  shaft  j,  (a)  when  friction  is  neglected,  and  (6) 
when  the  friction  of  the  gearing  is  considered. 

Friction  Neglected.  —  Neglecting  friction  is  equivalent  to  as- 
suming that  all  the  energy  supplied  to  the  shaft  a  is  transmitted 
to  the  shaft  j  and,  therefore,  the  work  of  the  power  P  in  one 
minute  must  be  equal  to  the  work  of  the  resistance  R  in  one 
minute.  Since  the  number  of  revolutions  per  minute  of  a  is 
Na  the  path  of  the  power  in  one  minute  is  Na  X  2  irp  and  the 
work  performed  by  it  in  one  minute  is  P  x  Na  X  2  irp;  and, 


208  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

similarly,  the  work  performed  by  the  resistance  in  one  minute  is 
R  x  N} X  2  irr,  whence 

P  x  Na  x  2  irp  =  R  x  NJ ;  x  2  TTT 

P  =  R*1PXK  <3> 

N- 
and  replacing  the  ratio  W  in  equation   (3)   by  its  value  from 

J\0 

equation  (2) 

P   —    7?    V  —   V   -^°X    -t  c   X    -/  e   X    -tgX    it  /  *x 

*  p  x  T6  x  Td  x  Tf  x  TA  x  T, 

and  we  may  write  that  the  power  required  to  overcome  a  resistance 
through  any  given  wheel  train,  when  friction  is  neglected,  is  equal  to 
the  resistance  multiplied  by  the  ratio  of  the  lever  arms  of  the  resistance 
and  the  power  with  respect  to  the  axes  of  the  shafts  of  the  last  and 
first  wheels,  respectively,  of  the  train,  multiplied  by  the  product  of 
the  numbers  of  the  teeth  of  all  the  drivers  in  the  train  divided  by  the 
product  of  the  numbers  of  the  teeth  of  all  the  followers. 

Friction  of  Gearing  Considered.  —  When  energy  is  transmitted 
from  one  shaft  to  another  through  a  wheel  train,  part  of  it  is 
lost  in  performing  wasteful  work  due  to  the  friction  between  the 
teeth  of  the  gears  and  that  between  the  shafts  and  their  bear- 
ings. To  determine  the  latter  it  is  necessary  to  know  the  normal 
pressures  between  all  the  shafts  and  their  bearings;  the  radii  and 
angular  velocities  of  all  the  shafts;  and  the  coefficients  of  friction, 
which  vary  with  the  normal  pressures,  the  speeds  of  the  rubbing 
surfaces,  the  materials  in  contact,  the  nature  of  the  lubricant,  if 
any,  and  the  temperature.  While  the  energy  lost  by  friction  be- 
tween the  shafts  and  their  bearings  can  be  taken  into  account 
with  reasonable  accuracy  in  any  particular  case,  it  will  not  be 
considered  in  this  discussion  as  it  is  ordinarily  considerably  less 
than  that  lost  through  the  friction  of  the  gearing. 

By  analytical  methods  and  by  experiment  the  percentage  of 
energy  transmitted  to  energy  received  has  been  determined  for 
a  pair  of  each  of  the  different  classes  of  gears.  These  percentages 
are  called  efficiencies  and  they  are  always  fractions  less  than  unity. 
Represent  the  efficiencies  of 

spur  gears  by    Et,  spiral  gears  by  Etp,  and 

bevel  gears  by  Eb,  worm-gears  by  Ew. 


TOOTHED  GEARING  209 

Now  since  the  velocity  ratio  of  the  shafts  a  and  j  is  constant  and 
not  affected  by  friction,  the  loss  of  energy  must  appear  entirely 
in  the  reduction  of  the  value  of  the  resistance  R  and,  therefore, 
the  value  of  P  that  will  produce  a  given  value  of  R  must  be  the 
theoretical  value  given  by  equation  (4)  divided  by  the  product 
of  the  efficiencies  of  all  the  pairs  of  gears  in  the  wheel  train, 
whence 


r      T  xT 

P  _  r>  v  j_   v    •LaKJ-c-e-g--t   v   _  -  _          /r\ 

X  p  X  TbxTdxT,xThxTi  x  EaxEbxE.pxEwxE, 

It  should  be  noted  that  this  equation  is  true  only  when  R  is  the 
resistance  and  P  the  driving  force.  If  P  becomes  the  resistance 
and  R  the  driving  force,  the  term  involving  the  efficiencies  must 
be  inverted  since  in  this  case  it  is  P  instead  of  R  that  must  be 
multiplied  by  this  term. 

103.  Efficiencies  of  Various  Classes  of  Gears.  —  The  efficiency 
of  a  pair  of  spur  gears  varies  from  about  .90  at  very  slow  speeds 
to  about  .985  when  the  linear  speed  of  a  point  on  the  pitch  cir- 
cumference is  equal  to  or  greater  than  200  feet  per  minute. 

When  the  linear  speed  of  a  toothed  gear  is  referred  to  it  is  to 
be  understood  as  meaning  the  speed  of  a  point  on  its  pitch  cir- 
cumference. 

The  efficiency  of  a  pair  of  spiral  or  worm-gears  varies  with  the 
helix  angle  which  the  thread  makes  with  a  line  perpendicular  to 
the  axis  of  the  gear  having  the  smaller  helix  angle,  and  also  with 
the  linear  speed  of  a  point  on  the  pitch  circumference.  At  very 
slow  speeds  the  efficiency  varies  from  about  .30  for  a  helix  angle 
of  five  degrees  to  about  .75  for  a  helix  angle  of  forty-five  degrees. 
At  speeds  equal  to  or  greater  than  200  feet  per  minute  the  efficiency 
varies  from  about  .75  for  a  helix  angle  of  five  degrees  to  about 
.90  for  a  helix  angle  of  forty-five  degrees.  The  low  efficiency  of 
spiral  and  worm-gearing  is  largely  due  to  the  thrusts  of  the  helical 
threads  in  directions  parallel  to  the  axes  of  the  shafts,  which 
develop  great  pressures  against  the  end  bearings  that  are  re- 
quired to  prevent  longitudinal  movement  of  the  shafts.  It  is 
evident  from  this  why  the  efficiency  increases  with  the  helix 
angle  of  the  threads.  By  interposing  ball  or  roller  bearings  be- 
tween the  shafts  and  their  end  bearings  the  efficiency  of  spiral 
and  worm-gearing  is  much  increased. 


210 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


The  efficiency  of  a  pair  of  bevel  gears  varies  from  about  .85 
at  very  slow  speeds  to  about  .95  at  speeds  equal  to  or  greater 
than  200  feet  per  minute. 

By  reference  to  equation  (5)  it  is  seen  how  great  is  the  in- 
crease required  in  the  value  of  the  power  over  its  theoretical 
value  to  produce  a  given  value  of  the  resistance  when  the  number 
of  gears  in  the  train  is  large  or  when  one  or  more  of  the  pairs 
consist  of  screw  gears.  Screw  gears  are  valuable,  however, 
when  it  is  desired  to  transmit  motion  between  nonparallel  shafts 
in  different  planes,  and  worm-gears  are  particularly  useful  when 
a  large  velocity  ratio  between  shafts  is  required. 

104.  Example.  —  The  elevating  mechanism  of  the  5-inch 
barbette  carriage,  model  of  1903,  is  shown  diagrammatically  in 
Fig.  87. 


Fig.  87. 

A  hand-wheel  a  with  crank-handle  j  is  mounted  on  a  short 
horizontal  shaft  carried  in  bearings  in  the  elevating  bracket. 
The  elevating  bracket  is  bolted  to  the  platform  bracket  and  the 
latter  is  bolted  to  the  pivot  yoke.  On  the  shaft  to  which  the 
hand-wheel  a  is  attached  is  a  bevel  gear  b  engaging  with  another 
c  on  an  inclined  shaft  d  also  carried  in  bearings  in  the  elevating 
bracket.  On  the  other  end  of  this  shaft  is  a  worm  e  engaging 
with  a  worm-wheel  /,  on  the  same  short  horizontal  shaft  g  with 
which  is  the  spur  gear  or  pinion  h  that  meshes  with  the  circular 
rack  i.  The  shaft  g  is  carried  in  bearings  in  the  elevating  bracket 
and  the  rack  i  is  attached  to  the  cradle  carrying  the  gun.  The 


TOOTHED  GEARING  211 

circular  rack  i  forms  a  part  of  a  spur  wheel  with  internal  teeth 
whose  axis  is  the  axis  of  the  trunnions  of  the  cradle.  The  trun- 
nion may,  therefore,  be  considered  as  the  shaft  to  which  i  is  at- 
tached. The  following  data  are  known  from  the  construction  of 
the  carriage,  viz.: 

Lever  arm  of  crank-handle  with  respect  to  axis  of  first  shaft 

Number  of  teeth  in  bevel  gear  b  =  18.  [  =  5  ins. 

Number  of  teeth  in  bevel  gear  c  =  30. 

Worm  is  single-threaded  (number  of  teeth  =  1). 

Number  of  teeth  in  worm-wheel  =  40. 

Number  of  teeth  in  spur  pinion  =  12. 

Number  of  teeth  in  spur  gear  of  which  rack  is  a  part  =  175. 

Radius  of  trunnions  =  2.505  ins. 

Coefficient  of  friction  at  trunnions  =  .1. 

Weight  of  parts  resting  on  trunnions  (gun  and  cradle)  =  14900  Ibs. 

Efficiency  of  a  pair  of  bevel  gears  =  .85. 

Efficiency  of  a  pair  of  worm-gears  =  .35. 

Efficiency  of  pinion  and  rack  (spur  gears)  =  .90. 

Neglecting  the  friction  between  the  shafts  and  their  bearings, 
let  it  be  required  to  determine  the  force  P  that  must  be  exerted 
on  the  crank-handle  j  to  start  the  gun  in  elevation,  assuming 
that  the  center  of  mass  of  the  parts  resting  on  the  trunnions  is 
in  the  axis  of  the  trunnions  so  that  the  weight  has  no  moment 
with  respect  to  that  axis. 

Solution.  —  The  resistance  to  be  overcome  is  the  force  of 
friction  at  the  cylindrical  surfaces  of  the  trunnions,  which  is 
equal  to  14900  x  .1  =  1490  Ibs.  This  resistance  has  a  lever 
arm  of  2.505  ins.  with  respect  to  the  axis  of  the  trunnions,  and 
the  lever  arm  of  the  power  with  respect  to  the  axis  of  the  first 
shaft  is  5  ins.  Therefore,  noting  from  Fig.  87  that  the  gears  6, 
e,  and  h  are  drivers  and  the  gears  c,  /,  and  i  are  followers,  we 
have,  in  accordance  with  the  principles  discussed  in  deducing 
equation  (5), 

,2.505        18x1x12        1_         _287ibs     (6) 

5  X  ~ 5~  X  30x40x175  X  .85x.35x.90  " 

2.87  Ibs.  is,  therefore,  the  force  on  the  crank-handle  required  to 
start  the  gun  in  elevation  or,  assuming  that  the  coefficient  of 


212  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

friction  at  the  trunnions  is  constant,  it  is  the  force  required 
to  keep  the  gun  moving  in  elevation  with  a  uniform  angular 
velocity. 

Force  on  the  Crank- Handle  Required  to  Produce  in  a  given 
Time  a  given  Angular  Velocity  of  the  Rotating  Parts.  —  To  at- 
tain a  given  angular  velocity  starting  from  rest  it  is  necessary, 
however,  to  give  the  rotating  parts  an  angular  acceleration  until 
the  given  angular  velocity  is  reached.  To  determine  the  in- 
crease in  the  force  required  on  the  crank-handle  to  give  the  gun 
a  desired  angular  velocity  in  elevation  in  a  given  time  is  not  a 
difficult  problem  providing  we  know  the  moment  of  inertia  of 
the  rotating  parts  about  the  axis  of  the  trunnions.  Having  de- 
cided on  the  angular  velocity  desired  and  the  time  in  which  it 
is  to  be  attained,  the  angular  acceleration  is  found  by  dividing  the 
angular  velocity  by  the  time.  To  produce  an  angular  accelera- 
tion d?<j>/dt2  about  a  fixed  axis  requires  a  moment  in  foot  pounds 
about  that  axis  equal  to  Sw2  x  d?<f>/dfi,  in  which  Smr2  is  the 
moment  of  inertia  in  mass  units  and  feet  of  the  rotating  parts 
about  the  axis.  While  the  force  required  to  produce  this  angular 
acceleration  is  actually  applied  at  the  pitch  circumference  of  the 
rack  fastened  to  the  cradle,  it  is  immaterial  where  it  be  taken 
as  applied  provided  its  moment  with  respect  to  the  axis  of  the 
trunnions  is  Smr2  x  d^/dP.  For  convenience  assume  the  force 
to  be  applied  at  the  surface  of  the  trunnions  as  is  the  friction  due 
to  the  weight  of  the  parts.  Then,  the  radius  of  the  trunnions 
being  2.505  ins.,  the  force  is 

Swr2  x  d^/dP ., 
2.505/12 

If  now  the  value  of  the  resistance,  1490  Ibs.,  in  equation  (6)  be 
increased  by  the  value  of  the  force  just  determined,  the  resulting 
value  of  P  will  be  the  force  on  the  crank-handle  required  to  over- 
come the  resistance  of  the  friction  at  the  trunnions  and  to  give 
the  gun  and  other  rotating  parts  the  desired  angular  velocity  in 
elevation  in  a  given  time.  Computations  such  as  these  are  im- 
portant in  designing  the  gearing  of  a  gun  carriage,  for  it  is  neces- 
sary that  the  gun  can  be  elevated  or  traversed  through  a  given 
number  of  degrees  in  a  reasonably  short  time  without  too  much 
effort  being  required  at  the  crank-handles. 


TOOTHED  GEARING  213 

105.  Pressure  between  the  Teeth  of  any  Pair  of  Gears  in  a 
Wheel  Train.  —  Although  it  has  been  shown  that  the  action  line 
of  the  pressure  between  the  teeth  when  the  involute  system  is 
used  makes  an  angle  of  15°,  and  when  the  cycloidal  system  is 
used  an  angle  varying  from  30°  to  0°,  with  the  common  tangent 
to  the  pitch  circles,  no  serious  error  will  be  committed  by  assum- 
ing it  to  be  coincident  with  this  tangent.  Referring  to  Fig.  85, 
the  pressure  between  the  teeth  of  any  two  gears  as  C  and  D  can 
be  determined  if  either  the  power  P  or  the  resistance  R  and  the 
corresponding  lever  arm  are  known.  If  the  resistance  R  and  its 
lever  arm  are  known  the  pressure  between  the  teeth  of  C  and  D 
may  be  regarded  as  a  power  with  respect  to  the  rest  of  the  train 
and  determined  as  was  the  power  P  except  that  the  part  of  the 
train  between  A  and  D  must  not  be  considered.  The  gears  C 
and  D  being  bevel  gears,  their  pitch  surfaces  are  conical  and  the 
lever  arm  of  the  pressure  between  the  teeth  with  respect  to 
the  axis  of  either  should  be  taken  equal  to  the  pitch  radius  at  the 
middle  of  the  length  of  the  teeth.  (In  practice  this  length  is 
called  the  face  of  the  teeth.)  If  the  power  P  and  its  lever  arm 
are  known  the  pressure  between  the  teeth  of  C  and  D  can  be  re- 
garded as  a  resistance  and  its  value  determined  by  considering 
only  that  portion  of  the  wheel  train  between  A  and  D. 

Frequently  only  the  horse-power  delivered  to  the  first  or  last 
shaft  of  a  wheel  train  and  the  speed  of  the  shaft  are  known. 
Horse-power  is  a  measure  of  the  rate  of  doing  work  and  one 
horse-power  is  equal  to  33000  ft.  Ibs.  per  min.  Let  the  number 
of  horse-power  delivered  to  the  shaft  a,  Fig.  85,  be  represented  by 
H  and  the  number  of  revolutions  per  minute  of  the  shaft  by  Na; 
then  if  the  point  of  application  of  the  force  required  to  produce 
this  number  of  horse-power  be  at  a  distance  of  p  inches  from  the 
axis  of  the  shaft,  the  path  it  will  follow  in  one  minute  is  2  irpNa/l2 
ft.,  and  the  intensity  of  the  force  required  to  produce  H  horse- 
power when  acting  with  a  lever  arm  of  p  inches  is 

H  x  33000  x  12 

2wpNa 

The  moment  of  this  force  (the  power)  with  respect  to  the  axis 

of  the  shaft  is,  in  in.  Ibs., 

H  x  33000  X  12  ~ 


214  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

Similarly,  if  the  horse-power  delivered  to  the  shaft  j  be  repre- 
sented by  H  and  the  number  of  revolutions  per  minute  of  the 
shaft  by  Nit  the  moment  of  the  resistance  in  in.  Ibs.  is 

H  x  33000  x  12 
Rr=~    -2^?T  (8> 

Having  obtained  either  the  moment  of  the  power  from  equa- 
tion (7)  or  the  moment  of  the  resistance  from  equation  (8),  the 
moment  of  the  other  can  be  obtained  from  equation  (5),  or  more 
readily  from  equation  (9)  below,  which  is  equation  (5)  put  in  a 
more  convenient  form  for  this  purpose. 

T  T  T  T  T  •  1 

Pv  -  Rr  x  °    l  V  (W 

*  TbTdTfThT,  x  EsEbE8pEwE, 

If  the  moment  of  the  power  is  known  the  moment  of  the  pres- 
sure between  the  teeth  of  any  pair  of  gears  in  the  train  can  be 
obtained  by  regarding  it  as  the  moment1- of  the  resistance  and 
considering  only  the  part  of  the  train  between  the  point  of  ap- 
plication of  the  power  and  that  of  the  pressure  between  the  teeth. 
If  the  moment  of  the  resistance  is  known  the  moment  of  the 
pressure  between  the  teeth  of  any  pair  of  gears  in  the  train  can 
be  obtained  by  regarding  it  as  the  moment  of  the  power  and  con- 
sidering only  the  part  of  the  train  between  the  point  of  applica- 
tion of  the  resistance  and  that  of  the  pressure  between  the  teeth. 
The  pressure  between  the  teeth  of  the  gears  can  then  be  obtained 
by  dividing  the  moment  of  the  pressure  by  the  pitch  radius  of 
the  driving  gear  in  case  the  pressure  is  regarded  as  a  resistance, 
and  by  the  pitch  radius  of  the  driven  gear  in  case  the  pressure  is 
regarded  as  a  power. 

106.  Stresses  in  the  Shafts.  —  The  pressure  between  the  teeth 
of  a  pair  of  gear-wheels  causes  a  torsional  moment  on  each  shaft 
equal  to  the  pressure  multiplied  by  the  pitch  radius  of  the  gear 
on  that  shaft.*  This  pressure  also  causes  a  bending  moment  and 
a  shearing  stress  in  each  shaft.  The  bending  moment  on  each 
shaft  will  depend  on  the  pressure  and  the  distance  between  the 
supports  (called  the  bearings)  of  the  shaft.  When  the  torsional 
and  bending  moments  and  the  diameters  of  the  shafts  are  known, 


*  Assuming  that  the  action  line  of  the  pressure  between  the  teeth  is  tangent 
to  the  pitch  circles  of  both  gears. 


TOOTHED  GEARING  215 

the  stresses  can  be  computed  as  explained  in  Chapter  IV.  The 
torsional  and  bending  stresses  should  be  combined  by  equations 
(24)  and  (25),  Chapter  IV. 

107.  Stresses  in  the  Gear-Teeth.  —  The  teeth  of  gears  may 
be  considered  as  cantilevers  subjected  to  a  bending  force  equal 
to  the  pressure  between  the  teeth.  In  practice  the  height  of  a 
tooth  above  the  pitch  circumference  is  made  equal  to  tjie  cir- 
cular pitch  divided  by  TT.  The  depth  below  the  pitch  circumfer- 
ence is  equal  to  the  height  above  it  increased  by  a  clearance 
equal  to  .05  times  the  circular  pitch,  the  clearance  being  allowed 
to  permit  the  rounding  of  the  corners  at  the  bottoms  of  the  tooth 
spaces  and  to  prevent  the  bottoming  of  the  teeth  in  the  tooth 
spaces.  The  thickness  of  the  tooth  at  the  pitch  circumference  is 
equal  to  one-half  the  circular  pitch.  The  thickness  at  the  root 
varies  with  the  circular  pitch  and  also  with  the  size  of  the  gear. 
Special  formulas  have  been  devised  for  determining  the  stresses  in 
gear-teeth  based  upon  their  variation  of  form  and,  consequently,  of 
the  variation  in  position  of  the  weakest  section,  which  generally 
occurs  a  little  above  the  root  of  the  tooth.  These  formulas  are 
given  in  mechanical  engineering  hand-books  and  in  works  on 
machine  design.  One  of  the  best  is  Lewis's  formula  which  is  as 
follows: 

W 


in  which  S  is  the  maximum  bending  stress;  W  is  the  pressure 
between  the  teeth  in  pounds;  p  is  the  circular  pitch  in  inches; 
/  is  the  face  of  the  teeth  in  inches,  that  is,  the  length  of  the  teeth 
measured  in  a  direction  parallel  to  the  axis  of  spur  gears,  and 
along  the  element  of  contact  in  bevel  gears;  and  n  is  the  number 
of  teeth  in  the  gear.  This  formula  will  answer  for  spur  gears, 
spiral  gears,  and  worm-wheels.  The  tooth  of  the  worm  is  always 
stronger  than  that  of  the  worm-wheel  with  which  it  meshes. 
For  bevel  gears  Lewis's  formula  is  as  follows: 

a  W  D 

\X  d 


216  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

in  which  S,  W,  f,  and  n  have  the  same  meanings  as  before,  p  is 
the  circular  pitch  in  inches  at  the  large  end  of  the  bevel-gear 
tooth,  D  is  the  pitch  diameter  at  the  large  end  of  the  tooth,  and 
d  is  the  pitch  diameter  at  the  small  end.  To  make  allowance  for 
shocks  when  the  speed  of  the  gearing  is  increased  the  factors  of 
safety  should  increase  with  the  speed. 

In  the  absence  of  a  special  formula  the  maximum  stress  in 
gear-teeth  may  be  approximately  obtained  with  the  error  on  the 
safe  side  by  considering  the  whole  pressure  uniformly  distributed 
along  the  top  edge  of  the  tooth  and  at  right  angles  to  the  plane 
passing  through  the  center  of  the  tooth  and  the  axis  of  the  wheel, 
by  taking  the  length  of  the  cantilever  as  the  total  height  of  the 
tooth  including  the  clearance,  and  by  considering  the  dangerous 
section  as  located  at  the  root  and  as  having  a  thickness  equal  to 
that  of  the  teeth  at  the  pitch  circumference,  that  is,  equal  to  one- 
half  the  circular  pitch.  The  height  of  bevel-gear  teeth  should  be 
taken  equal  to  half  the  sum  of  the  heights  at  the  large  and  small 
ends,  and  the  thickness  as  half  the  sum  of  the  thicknesses  at  the 
large  and  small  ends. 

108.  Stresses  in  the  Arms  of  Gear- Wheels.  —  The  teeth  of 
large  gears,  or  gear-wheels,  are  formed  on  a  rim  connected  to  a 
nave  or  hub  by  several  arms.  The  stresses  in  the  arms  may  be 
determined  by  considering  them  as  cantilevers  subjected  to  a 
bending  moment  equal  to  the  pressure  between  the  teeth  multi- 
plied by  the  length  of  the  arm  divided  by  the  number  of  arms. 
In  order  to  err  somewhat  on  the  safe  side  the  length  of  the  arm  is 
taken  equal  to  the  pitch  radius  of  the  gear. 

Proportions  for  the  Rims  of  Gear- Wheels.  —  Arbitrary  pro- 
portions for  the  rims  of  gear-wheels  based  on  experience  have 
been  adopted.  These  proportions  are  given  in  mechanical  en- 
gineering hand-books  and  standard  works  on  machine  design. 
A  standard  thickness  for  rims  of  gear-wheels  whose  circular  pitch 
is  greater  than  one  and  one-half  inches  is  one-half  the  circular 
pitch.  For  gear-wheels  having  pitches  less  than  one  and  one- 
half  inches  the  thickness  of  the  rim  is  taken  as 

.4  p  +  1/8  in.  (12) 

in  which  p  is  the  circular  pitch  in  inches.  A  rib  is  added  to  the 
under  side  of  the  rim.  The  width  of  the  rib  is  the  same  as  that 


TOOTHED  GEARING  217 

of  the  arms  of  the  gear-wheel  and  its  height  is  equal  to  the  thick- 
ness of  the  rim. 

Proportions  for  the  Hubs  of  Gear-  Wheels.  —  Arbitrary  pro- 
portions for  the  hubs  of  gear-wheels  adopted  as  a  result  of  ex- 
perience are  also  given  in  mechanical  engineering  hand-books 
and  standard  works  on  machine  design.  If  the  gear-wheel  has 
to  transmit  the  full  power  of  the  shaft  the  thickness  of  the  hub 
is  made  equal  to  the  radius  of  the  shaft.  If  the  gear-wheel  does 
not  have  to  transmit  the  full  power  of  the  shaft  the  thickness  of 
the  hub  is  given  by  the  following  formula: 


t  =       pR/3  (13) 

in  which  t  is  the  thickness  of  the  hub  in  inches,  /  is  the  face  of  the 
teeth  in  inches,  p  is  the  circular  pitch  in  inches,  and  R  is  the  pitch 
radius  in  inches.  The  length  of  the  hub  varies  from  /  to  1.4/. 

109.  Gear  Cutting.  —  The  teeth  of  spur  gears,  spiral  gears, 
and  worm-wheels  are  cut  in  a  universal  milling  machine  such  as 
that  which  the  cadets  have  operated  in  their  practical  work. 
When  a  large  amount  of  gear  cutting  has  to  be  done  special 
milling  machines  are  used  which  are  peculiarly  adapted  to  the 
work  and  are  not  capable  of  performing  the  miscellaneous  char- 
acter of  work  done  by  the  regular  milling  machine.  The  par- 
ticular feature  of  the  special  machines  which  renders  them  most 
useful  is  that  they  are  automatic  in  their  action.  After  being 
started  they  continue  to  work  without  further  attention  until  all 
the  teeth  are  cut.  After  one  tooth  space  has  been  cut  the  gear- 
wheel is  moved  past  the  cutter  and  rotated  through  the  proper 
angle  so  as  to  be  ready  for  the  next  cut,  all  by  the  automatic  oper- 
ation of  the  machinery.  As  the  principles  of  the  automatic  ma- 
chines are  the  same  as  those  of  the  milling  machine  with  which 
the  cadets  are  familiar,  the  latter  will  be  referred  to  in  the^  de- 
scriptions that  follow. 

Rotary  Cutters.  —  The  profiles  of  the  teeth  of  the  milling 
cutters  used  are  exact  duplicates  of  those  of  the  tooth  spaces 
they  are  required  to  produce.  The  manufacture  of  milling  cut- 
ters is  a  specialty  not  much  practised  by  the  ordinary  machine 
shop  for,  by  reason  of  the  special  appliances  used  and  the  quantity 
made  at  one  time,  the  manufacturers  of  the  cutters  can  furnish 
a  better  article  for  less  money  than  it  would  cost  to  make  it  in  an 


218 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


ordinary  shop.  A  brief  description  of  the  general  method  of 
making  cutters  for  gear-teeth  is  as  follows:  A  piece  of  tool  steel 
is  turned  in  a  lathe  to  the  proper  diameter  and  width  and  then 
notches  for  the  teeth  are  cut  in  the  milling  machine.  The  piece 
is  then  taken  back  to  the  lathe  and  the  outline  of  the  teeth  cut  by 
a  lathe  tool,  shown  in  Fig.  88,  the  cutting  edges  of  which  have  the 
outline  of  a  tooth  space  and  have  been  shaped  by  filing  to  a 
template  prepared  from  a  drawing  of  the  tooth  curves.  As  the 
top  surfaces  of  the  teeth  of  a  milling  cutter  are  not  concentric 
with  its  axis  but  are  of  a  spiral  character  to  give  proper  clearance 
and  keenness  to  the  cutting  edges,  as  shown  by  the  lines  ab  in 
Fig.  88  in  which  the  points  a  are  at  the  cutting  edges,  the  lathe 


Lathe  Tool. 


Fig.  88. 


Rotary  Cutter. 


is  provided  with  an  attachment  called  a  backing  off  or  relieving 
attachment  which  forces  the  lathe  tool  gradually  in  as  the  work 
is  revolving  past  it  from  a  to  6  and  then  draws  it  out  quickly  to 
be  in  position  to  cut  the  edge  a  at  the  proper  height  on  the  next 
tooth.  After  the  outlines  of  the  teeth  have  been  cut  in  the  lathe 
the  cutter  is  hardened  and  tempered. 

110.  Cutting  the  Teeth  of  Spur  Gears.  —  The  wheel  or  gear 
before  the  teeth  are  cut  is  called  a  gear-blank.  This  is  turned 
to  the  required  shape  and  size  in  the  lathe.  It  is  then  mounted 
on  centers  between  the  index  head  and  tail  stock  of  the  milling 


TOOTHED  GEARING  219 

machine.  The  rotary  cutter  is  placed  on  the  arbor  with  its 
center  in  the  vertical  plane  containing  the  axis  of  the  gear-blank, 
and  the  table  raised  to  give  the  proper  depth  of  cut.  The  ma- 
chine is  then  started,  the  feed  mechanism  thrown  in,  and  the 
table  fed  past  the  cutter.  It  is  then  drawn  back  by  hand,  the 
blank  turned  through  the  proper  angle  by  the  index  mechanism, 
and  another  cut  taken  and  so  on. 

Cutting  the  Teeth  of  Spiral  Gears.  —  The  method  of  cutting 
the  teeth  of  spiral  gears  is  essentially  the  same  in  principle  as 
for  spur  gears,  the  only  difference  being  that  the  teeth  are  helical. 
The  table  of  the  milling  machine  is,  therefore,  set  at  the  proper 
angle  with  the  axis  of  the  spindle,  and  the  index  head  is  con- 
nected by  the  requisite  gearing  to  the  feed  screw  of  the  table  in 
order  to  give  the  gear-blank  such  a  combined  motion  of  trans- 
lation and  rotation  as  will  secure  the  proper  helix  angle,  these 
operations  being  similar  to  those  performed  by  the  cadets  in  cut- 
ting the  spiral  teeth  of  a  reamer. 

111.  Cutting  the  Teeth  of  a  Worm  and  Worm- Wheel.  —  The 
worm  is  turned  and  threaded  in  the  lathe,  the  thread  being  cut 
in  practically  the  same  way  as  the  thread  of  an  ordinary  bolt 
except  that  a  roughing  tool  is  first  used,  being  followed  by  the 
finishing  tool.  The  roughing  tool  has  a  square  nose  slightly 
narrower  than  the  tooth  space  at  the  bottom  of  the  thread. 
When  the  diameter  at  the  bottom  of  the  thread  has  been  re- 
duced to  the  prescribed  size  with  the  roughing  tool,  the  finish- 
ing tool  is  put  in  the  tool  post  and  the  thread  finished  with  it. 
The  finishing  tool  is  very  carefully  made  and  filed  to  a  template 
whose  profile  is  the  same  as  that  of  the  tooth  space  of  the 
worm. 

In  order  that  the  teeth  of  a  worm  and  worm-wheel  whose  axes 
are  perpendicular  to  each  other  shall  be  in  contact  over  a  con- 
siderable arc  measured  on  the  pitch  surface  of  the  worm  in  a 
direction  parallel  to  its  thread,  the  face  of  the  wheel  is  made  con- 
cave to  fit  the  curvature  of  the  worm  and  the  teeth  are  cut  in  a 
milling  machine  using  a  rotary  cutter  called  a  hob.  The  worm- 
wheel  blank  is  turned  in  a  lathe.  If  the  number  of  worm-wheels 
under  manufacture  justifies  the  expense,  particularly  if  the 
wheels  are  large,  an  attachment  is  made  or  purchased  for  turning 
the  concave  face  of  the  wheel.  This  attachment  is  placed  on  the 


220  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

tool  carriage  of  the  lathe  and  permits  the  cutting  tool  to  be  fed 
around  a  fixed  center  on  the  carriage,  thus  turning  the  concave 
face  to  the  required  radius.  The  longitudinal  feed  of  the  lathe 
is  not  used  with  this  attachment.  The  cross  feed  is  used  only  to 
obtain  the  proper  depth  of  cut.  If  only  a  few  worm-wheels  are 
to  be  made,  particularly  if  the  wheels  are  small,  an  attachment 
such  as  described  would  not  be  used.  Instead  the  concave  face 
of  the  wheel  is  first  roughed  out  with  a  round-nose  tool  to  ap- 
proximately the  correct  outline  as  shown  by  a  template,  using 
the  power  longitudinal  feed  and  operating  the  cross  feed  by 
hand.  The  cut  is  started  at  the  right  edge  of  the  face  of  the 
wheel  and  the  tool  is  fed  to  the  left  by  power  while  being  fed  in 
by  hand  in  such  manner  as  to  make  it  follow  approximately  the 
arc  of  a  circle.  When  the  tool  reaches  the  center  of  the  face 
it  is  withdrawn  from  the  work  and  a  new  cut  is  started  at  the 
left  edge  of  the  face,  the  tool  now  being  fed  to  the  right  by  power 
while  being  fed  in  by  hand  as  before.  When  the  second  cut 
reaches  the  first  at  the  center  of  the  face  the  tool  is  again  with- 
drawn from  the  work,  and  the  operations  just  described  are  re- 
peated as  often  as  may  be  necessary  until  the  face  of  the  wheel  is 
rough  turned  to  approximately  the  correct  size  and  shape.  The 
face  is  then  finished  to  the  template  using  a  broad-nose  finishing 
tool  whose  cutting  edge  is  rounded  to  the  same  radius  as  the 
template.  This  tool  is  ordinarily  narrower  than  the  face  of  the 
wheel  and  is  only  used  to  dress  off  the  high  spots  indicated  by 
the  template,  being  shifted  from  one  spot  to  another  of  the  face 
of  the  wheel  for  this  purpose.  When  the  finishing  tool  is  used 
all  feeding  is  done  by  hand. 

The  hob,  see  Fig.  89,  is  made  of  tool  steel  and  is  almost  the 
exact  counterpart  of  the  worm,  the  only  differences  being  that 
its  diameter  is  slightly  greater  to  provide  for  the  clearances  at 
the  bottoms  of  the  tooth  spaces  of  the  wheel,  that  it  is  fluted  in 
a  direction  parallel  to  its  axis  so  as  to  form  cutting  teeth,  and 
that  the  top  surfaces  of  the  teeth  like  those  of  the  milling  cutter 
referred  to  in  article  109  are  of  a  spiral  form  to  give  proper  clear- 
ance and  keenness  to  the  cutting  edges.  It  is  turned  and  threaded 
in  the  lathe  and  fluted  in  the  milling  machine.  The  top  surfaces 
of  the  teeth  are  then  filed  to  give  them  the  proper  shape  and  the 
hob  is  finally  hardened  and  tempered. 


ENGINEERING  BUREAU 
QAMMOK 


TOOTHED  GEARING  221 

To  cut  the  teeth  of  the  worm-wheel  it  is  mounted  on  an  arbor 
between  the  index  centers  of  the  milling  machine  and  left  free  to 
rotate  thereon.  It  is  then  brought  directly  under  the  hob  which 
is  fastened  to  an  arbor  in  the  spindle  of  the  machine,  the  axes 
of  the  hob  and  worm-wheel  being  perpendicular  to  each  other. 
The  machine  is  started  and  the  table  raised  until  the  hob  teeth 
sink  a  certain  distance  into  the  face  of  the  wheel.  The  rotation 
of  the  hob  while  cutting  will,  because  of  its  cutting  threads, 
cause  the  wheel  to  rotate  in  the  same  way  as  if  it  were  in  mesh 
with  its  worm.  After  the  wheel  has  made  one  complete  revolu- 
tion the  hob  is  sunk  deeper  into  it  and  another  cut  taken  on  the 
teeth  all  around  the  circumference,  these  operations  being  re- 
peated until  the  tooth  spaces  in  the  wheel  are  cut  to  the  proper 
depth.  Very  frequently,  and  especially  if  the  wheel  is  large,  it 
is  "gashed"  by  an  ordinary  milling  cutter  before  being  hobbed 
as  when  this  is  done  there  is  no  danger  of  the  wheel  not  rotating 
properly  with  the  hob,  which  sometimes  occurs  when  the  wheel 
is  not  gashed.  A  gash  is  cut  over  every  tooth  space,  the  table 
of  the  machine  being  turned  so  that  the  gashes  are  parallel  to 
the  helices  of  the  finished  teeth  at  the  central  plane  of  the  wheel. 
The  wheel  does  not  rotate  when  the  gashes  are  being  made  but 
it  is  turned  through  the  proper  angle  between  gashes  by  the  index 
mechanism. 

In  some  machines  the  worm-wheel  while  being  hobbed  is  not 
left  free  to  rotate  on  the  centers  but  is  required  by  gearing  to 
rotate  at  exactly  the  correct  speed  with  reference  to  that  of  the 
hob.  When  worm-wheels  are  hobbed  in  these  machines  the 
preliminary  gashing  is  not  needed. 

Fig.  89  shows  the  operation  of  hobbing  a  worm-wheel. 

When  the  axes  of  a  worm  and  worm-wheel  are  not  to  be  per- 
pendicular to  each  other,  the  wheel  is  not  hobbed  but  instead  its 
teeth  are  cut  with  an  ordinary  rotary  cutter  in  the  same  way  as 
the  teeth  of  a  spiral  gear. 

112.  Cutting  the  Teeth  of  Bevel  Gears.  —  Because  of  the 
conical  pitch  surfaces  of  bevel  gears  a  section  of  the  tooth  at  one 
end  is  larger  than  at  the  other,  and  the  tooth  outline  varies  in 
size  from  the  large  end  to  the  small  end;  consequently  perfectly 
formed  bevel-gear  teeth  cannot  be  cut  by  a  rotary  cutter.  Never- 
theless such  teeth  were  in  the  past  cut  in  the  milling  machine 


222 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


**? 
TO' 

oo 


w 

o 
cr 

cr 

5' 

CR 


TOOTHED  GEARING 


223 


with  a  rotary  cutter  and  this  practice  is  followed  to  a  considerable 
extent  at  the  present  time. 

After  the  gear-blank  has  been  turned  in  the  lathe,  if  the  teeth 
are  to  be  cut  in  a  milling  machine  it  is  mounted  on  an  arbor  in 
the  spindle  of  the  index  head,  which  is  rotated  around  a  hori- 
zontal axis  until  the  middle  element  of  the  surface  that  will  form 
the  bottom  of  the  tooth  space  is  in  a  horizontal  plane.  A  rotary 
cutter  whose  width  is  that  of  the  tooth  space  at  its  small  end  and 
the  profile  of  whose  cutting  edges  is  the  same  as  that  of  the  tooth 
space  at  its  large  end  is  so  mounted  on  an  arbor  in  the  spindle  of 
the  machine  that  its  center  and  the  axis  of  the  gear-blank  are 


Fig.  90. 

contained  in  the  same  vertical  plane,  and  two  spaces,  a  and  6, 
are  cut  as  shown  in  Fig.  90,  the  index  mechanism  being  used  to 
obtain  the  proper  spacing. 

This  will  leave  the  tooth  c  too  wide  at  the  large  end.  The 
table  of  the  milling  machine  is  then  moved  by  the  cross  feed  in 
the  direction  of  the  arrow  a  certain  distance  ed,  which  moves  the 
center  of  the  gear-blank  to  the  right  of  the  center  of  the  cutter. 
The  gear-blank  is  then  revolved  until  the  cutter  will  just  enter 
the  cut  a  at  the  small  end  of  what  will  be  the  finished  tooth  space. 
The  center  of  the  blank  being  now  to  the  right  of  the  center  of 


224 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


the  cutter,  the  path  of  the  latter  when  the  blank  is  fed  up  to  it 
will  not  be  parallel  to  the  original  cut  a,  but  will  make  an  angle 
with  it;  and  the  cutter  will  cut  the  left  side  of  the  tooth  c  as 
shown  by  the  dotted  line  fg,  the  amount  of  metal  removed  vary- 
ing from  practically  nothing  at  the  small  end  of  the  tooth  to  an 
amount  at  the  large  end  dependent  on  the  distance  ed  which  the 
center  of  the  gear-blank  has  been  moved  to  the  right  of  the  center 
of  the  cutter.  The  center  of  the  blank  is  next  moved  as  far  to 
the  left  of  the  center  of  the  cutter  as  it  was  to  the  right  and, 
after  the  blank  is  revolved  until  the  cutter  will  enter  6  at  the 
small  end  of  which  will  be  the  finished  tooth  space,  a  cut  is  taken 
on  the  right  side  of  c  as  shown  by  the  dotted  line  hi.  The  thick- 
ness of  c  at  the  pitch  circumference  at  the  large  end  is  then  meas- 
ured; if  too  thick  the  operations  just  described  are  repeated,  the 
center  of  the  blank  being  set  a  little  further  to  the  right  and  to 
the  left  of  the  center  of  the  cutter  than  it  was  before.  When  the 
correct  thickness  of  the  tooth  at  the  pitch  circumference  at  the 
large  end  is  obtained,  cuts  are  taken  all  around  the  blank  with  its 
center  set  at  the  proper  distance  to  the  right  of  the  center  of  the 
cutter,  and  then  with  the  center  of  the  blank  set  at  the  same 
distance  to  the  left  of  the  center  of  the  cutter.  The  index  mechan- 
ism is  used  to  rotate  the  blank  through  an  angle  corresponding 
to  the  circular  pitch. 


c  a 


Before  filing. 


Fig.  91. 


After  filing. 


This  method  will  produce  a  proper  outline  at  the  large  end  of 
the  tooth  but  toward  the  small  end  it  will  not  be  curved  enough, 
as  will  be  seen  from  Fig.  91,  which  shows  the  large  end  of  a  bevel- 
gear  tooth  placed  over  the  small  end  of  the  same  tooth,  it  being 
understood  that  the  part  of  the  cutter  that  forms  the  portion  ab 
of  the  large  end  must  also  form  the  whole  of  the  side  cd  of  the 


TOOTHED  GEARING 


225 


small  end  of  the  tooth.  The  parts  of  the  teeth  near  their  small 
ends  must,  therefore,  be  rounded  with  a  file. 

113.  Planing  Bevel-Gear  Teeth.  — The  proper  way  to  cut 
bevel-gear  teeth  is  to  plane  them,  using  for  this  purpose  a  specially 
designed  shaper.  The  teeth  of  all  bevel  gears  for  gun  carriages 
are  required  to  be  planed. 

Since  the  tooth  surfaces  of  bevel  gears  are  generated  by  cones 
rolling  on  conical  pitch  surfaces  whose  apexes  are  at  the  point 
of  intersection  of  the  shafts,  or  by  planes  rolling  on  base  cones 
whose  apexes  are  at  the  same  point,  it  is  evident  that  the  elements 


CUTTING   TOOL 


CUTTING  TOOL 


Fig.  92. 


of  the  tooth  surfaces  are  straight  lines  converging  to  the  point  of 
intersection  of  the  shafts.  If,  therefore,  the  point  of  a  planing  tool 
can  be  made  to  follow  these  elements,  the  surfaces  cut  by  it  will 
be  correct  tooth  surfaces. 

If  one  end  of  a  rod  be  pivoted  at  a,  Fig.  92,  by  a  universal 
joint  and  the  other  end  be  moved  around  the  inside  of  a  semi- 
circle A,  it  is  apparent  that  it  will  by  its  motion  generate  the  half 
surface  of  a  cone,  and  that  any  section  of  this  half  cone  parallel 
to  the  plane  of  the  semi-circle  A  will  also  be  a  semi-circle  whose 
size  will  vary  with  the  distance  of  the  section  from  a.  If  the  end 


226  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

of  the  rod  instead  of  being  moved  continuously  around  the  semi- 
circle be  moved  around  it  step  by  step,  the  successive  positions 
of  the  rod  being  quite  close  to  each  other,  a  slide  carrying  a  cut- 
ting tool  can  be  placed  on  the  rod  and  made  to  move  forward 
toward  a  and  back  again  at  every  successive  position  of  the  rod, 
which  in  this  case  would  be  bent  downward  for  a  short  distance 
at  its  end  before  being  pivoted  at  a  in  order  to  enable  the  point 
of  the  cutting  tool  to  travel  directly  toward  a  and  back.  If  a 
metal  bar  not  too  long  is  placed  between  A  and  a  and  the  first 
position  of  the  free  end  of  the  rod  carrying  the  slide  and  cutting 
tool  is  at  6,  it  is  apparent  that,  during  the  successive  movements 
of  the  free  end  of  the  rod  around  the  semi-circle  from  6  to  c,  the 
cutting  tool  on  the  slide,  by  moving  forward  and  backward  each 
time  the  rod  takes  a  new  position,  will  cut  in  the  bar  a  hollow  half 
cone  whose  apex  is  at  a  and  whose  sections  parallel  to  the  plane 
of  A  will  all  be  semi-circles.*  If  instead  of  moving  the  free  end 
of  the  rod  around  the  inside  of  the  semi-circle  it  be  moved  around 
the  inside  of  the  rectangle  B  from  6  to  c,  the  tool  can  be  made  to 
cut  a  pyramidal  trough  whose  apex  is  at  a  in  a  metal  bar  placed 
between  B  and  a.  Any  section  of  this  trough  parallel  to  the 
plane  of  B  will  give  three  sides  of  a  rectangle  similar  to  B,  the 
size  of  the  rectangle  varying  with  the  distance  of  the  section 
from  a.  And  by  substituting  for  the  rectangle  a  template 
having  the  outline  of  a  correctly  shaped  tooth  space,  the  tool  can 
be  made  to  cut  a  correctly  shaped  tooth  space  in  a  bevel-gear 
blank  so  placed  between  C  and  a  that  the  apex  of  its  conical  pitch 
surface  is  at  a.  All  elements  of  the  tooth  surfaces  so  cut  will 
converge  toward  a  and  the  difference  in  the  size  of  the  tooth  at 
its  large  and  small  ends  will  depend  upon  the  length  of  its  face  and 
the  angle  at  the  apex  of  the  pitch  cone.  The  teeth  of  bevel  gears 
of  different  size  can  be  cut  by  varying  the  distance  of  the  gear- 
blank  from  the  point  a. 

The  construction  of  a  bevel-gear  shaper  is  somewhat  com- 
plicated but  the  principle  upon  which  it  operates  is  not  difficult 
to  understand  after  the  discussion  in  the  preceding  paragraph. 
The  ram  of  the  shaper  moves  in  guideways  in  a  saddle  pivoted 

*  To  permit  the  end  of  the  rod  to  be  fed  around  the  inside  of  the  semi-circle, 
rectangle,  or  template  while  the  tool  is  cutting  the  bar  (or  gear  blank),  the 
latter  would  ordinarily  have  to  be  first  roughed  out  by  central  cuts. 


TOOTHED  GEARING  227 

at  a  point  in  front  by  a  universal  joint.  The  rear  part  of  the 
saddle  carries  a  roller  which  the  mechanism  of  the  machine  keeps 
hard  pressed  against  a  template  whose  outline  is  that  of  a  cor- 
rectly shaped  tooth  space  on  a  large  scale.  The  ram  carries  the 
cutting  tool  and  is  moved  backward  and  forward  by  the  ordinary 
shaper  mechanism.  After  each  stroke  of  the  shaper  the  feed 
mechanism  moves  the  roller  at  the  rear  end  of  the  saddle  down- 
ward while  at  the  same  time  it  is  kept  hard  pressed  by  springs  or 
otherwise  against  the  tooth  outline  of  the  template.  In  this 
way  the  rear  part  of  the  saddle,  which  corresponds  to  the  rod 
in  Fig.  92,  follows  the  outline  of  a  correctly  shaped  tooth  space 
while  its  front  end  is  pivoted  at  a  fixed  point.  The  ram  of  the 
shaper  corresponds  to  the  slide  on  the  rod  of  Fig.  92.  The  bevel- 
gear  blank  on  which  the  teeth  are  to  be  cut  is  placed  in  front  of 
the  ram  with  the  apex  of  its  pitch  cone  at  the  point  to  which  the 
front  end  of  the  saddle  is  pivoted.  The  position  of  the  gear- 
blank  in  front  of  the  ram  can  be  varied  to  suit  the  size  of  gear 
being  cut.  After  one  tooth  space  has  been  cut  the  blank  is  auto- 
matically rotated  through  the  proper  angle  to  place  it  in  position 
for  another  cut  by  a  mechanism  similar  to  the  index  mechanism 
of  a  milling  machine.  In  this  machine  the  tool  would  be  fed 
downward  by  the  downward  movement  of  the  rear  end  of  the 
saddle  along  the  tooth  outline  of  the  template,  and  one  side  of  a 
tooth  would  be  cut  at  a  time. 


CHAPTER  VI. 
COUNTER-RECOIL   SPRINGS. 

114.  Helical  Springs.  —  The  counter-recoil  springs  used  in  gun 
carriages  for  returning   the  gun  into  battery  after   recoil  has 
ended  are  made  of  steel  bars  coiled  into  helices  and  hence  are 
called  helical  springs.     The  cross-section  of  the  bars  from  which 
the  springs  are  made  is  generally  either  circular  or  rectangular. 
The  steel  from  which  the  bars  are  forged  must  be  of  the  best 
quality,  and  since  an  exceptionally  high  elastic  limit  is  required 
the  carbon  content  of  the  steel  is  high  and  the  springs  are  hardened 
and  tempered  before  use;  and,  as  it  is  necessary  to  retain  as  much 
as  possible  of  the  increased  elastic  limit  caused  by  the  hardening 
process,  the  tempering  temperature  is  low. 

Fig.  93  shows  two  counter-recoil  springs,  one  coiled  from  a  bar 
of  circular  cross-section  and  the  other  coiled  from  a  bar  of  rec- 
tangular cross-section. 

In  order  that  each  end  of  a  counter-recoil  spring  shall  bear 
evenly  against  the  piston  of  the  spring  rod  or  other  part  against 
which  it  acts,  the  ends  of  the  end  coils  are  closed  down  against 
the  adjoining  coils  and  ground  flat  so  that  the  end  surfaces  of 
the  springs  are  truly  at  right  angles  to  its  axis.  When  this  is  the 
case  the  force  acting  on  the  spring  is  uniformly  distributed  over 
its  end  surface  and  the  effect  is  the  same  as  if  the  force  were  con- 
centrated at  the  axis  of  the  spring  and  acted  in  the  direction  of 
that  axis. 

115.  Torsional  Stresses  and  Strains  in  a  Straight  Bar.  —  As 
the  stresses  and  strains  produced  in  the  material  of  a  helical 
spring  by  a  force  acting  on  it  in  the  direction  of  its  axis  are  almost 
entirely  those  of  torsion,  the  effect  of  a  twisting  force  on  a  straight 
bar  will  be  considered  before  studying  the  stresses  and  strains 
produced  in  a  helical  spring  by  a  force  acting  to  compress  or 
elongate  it. 

228 


COUNTER-RECOIL  SPRINGS 


229 


Fig.  93.  —  Counter-Recoil  Springs. 


230  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

Let  BA,  Fig.  94,  be  a  bar  fixed  at  the  end  B  and  subjected  to 
a  torsional  moment  caused  by  the  force  F.  Then,  as  stated  in 
article  71,  page  111,  the  torsional  stress  per  unit  of  area  in  the 
extreme  fibre  of  any  section  of  the  bar  will  be 

rStf"  =S  =  Mtr/Ip  (1) 


Fig.  94. 
in  which 

St"  is  the  torsional  stress  per  unit  of  area  at  a  unit's  distance 
from  the  axis  of  the  bar, 

S  is  the  torsional  stress  per  unit  of  area  in  the  extreme  fibre  of 
a  section, 

r  is  the  distance  in  inches  from  the  axis  of  the  bar  to  the  ex- 
treme fiber  of  a  section, 

Mt  is  the  torsional  moment  in  in.  Ibs.,  and 

IP  is  the  polar  moment  of  inertia  of  the  section  in  ins.4 

In  Fig.  94,  which  is  reproduced  from  page  37,  Fiebeger's  Civil 
Engineering,  ef  is  a  surface  fibre  before  being  distorted  by  the 
twisting  force  F  and  ed  is  the  position  taken  by  this  fibre  when  the 
force  acts  on  the  bar.  Every  cross-section  of  the  bar  except  that 
at  the  fixed  end  B  is  rotated  around  the  axis  ba  by  the  twisting 
force,  each  section  being  rotated  through  a  slightly  greater  angle 


. 

COUNTER-RECOIL  SPRINGS  231 

than  the  one  next  it  on  the  side  of  the  fixed  end  of  the  bar.  The 
maximum  rotation  occurs  in  the  section  at  the  free  end,  whose 
plane  contains  the  force,  and  the  rotation  of  any  other  section 
is  equal  to  the  maximum  rotation  multiplied  by  the  distance  of 
the  section  under  consideration  from  the  fixed  end  of  the  bar 
divided  by  the  length  of  the  bar.  If  the  force  did  not  act  at 
the  end  of  the  bar  the  maximum  rotation  would  still  occur  in  the 
section  whose  plane  contains  the  force,  and  the  rotations  of 
the  other  sections  nearer  the  fixed  end  of  the  bar  would  bear  the 
same  relations  to  the  maximum  rotation  as  before,  providing  we 
substitute  for  the  length  of  the  bar  the  distance  from  the  section 
whose  plane  contains  the  force  to  the  fixed  end.  All  sections 
between  the  force  and  the  free  end  of  the  bar  would  have  the 
same  rotation  as  that  of  the  section  whose  plane  contains  the 
force. 

Let     L  be  the  length  of  the  bar  shown  in  Fig.  94, 
I,  the  length  of  the  arc  fd,  and 

E,  the  torsional  modulus  of  elasticity,  sometimes  called 
also  the  coefficient  of  torsional  elasticity; 

then,  as  shown  on  page  40,  Fiebeger's  Civil  Engineering, 


or  substituting  for  S  its  value  from  equation  (1)  and  solving  for 
l/L 


Since  I  is  the  length  of  the  arc  fd  and  r  is  the  distance  of  the 
extreme  fibre  from  the  axis  of  the  bar,  l/r  is  the  measure  of  the 
angle,  expressed  in  radians,  through  which  has  rotated  the  section 
whose  plane  contains  the  twisting  force.  Solving  equation  (3) 
for  l/r  we  have 


116.  Stresses  in  Helical  Springs.  —  Let  Fig.  95  represent  a 
helical  spring  resting  on  a  flat  surface  JJ  and  subjected  to  com- 
pression by  the  force  C.  Since  the  action  line  of  the  force  coin- 


232 


STRESSES  IN  GUNS  AND  GUN  CARRIAGES 


tides  with  the  axis  of  the  spring  it  is  symmetrically  placed  with 
respect  to  each  section  of  the  bar  from  which  the  spring  is  coiled, 
and  the  stresses  produced  by  it  in  all  sections  will  be  the  same. 
The  part  of  the  bar  between  the  point  of  application  of  the  force 


Fig.  95. 

and  section  a  is  not  considered  in  this  discussion  as  it  is  an  end 
coil  that  has  been  flattened  and  ground  to  provide  an  end  surface 
perpendicular  to  the  axis  of  the  spring.  End  coils  are  called 
ineffective  coils;  they  are  not  relied  upon  to  add  to  the  spring 
effect  as  a  whole  but  merely  to  transmit  the  force  to  the  rest  of 
the  spring. 

Let  a  be  a  right  section  of  the  bar.  As  the  bar  is  coiled  into 
a  helix  this  section  and  every  other  right  section  will  be  slightly 
inclined  from  the  vertical,  and  the  component  of  the  force  C 
perpendicular  to  the  section  will  cause  a  slight  bending  stress 
and  a  slight  compressive  stress  therein;  but  as  these  stresses  are 
small  they  are  neglected  in  spring  computations,  and  the  plane  of 


COUNTER-RECOIL  SPRINGS  233 

a  right  section  of  the  bar  is  assumed  to  be  parallel  to  the  axis  of 
the  spring  and  to  the  action  line  of  the  force.  Since  the  bar  is 
coiled  around  the  axis  of  the  spring,  the  plane  of  section  a  and  of 
every  other  right  section  of  the  bar,  considered  as  vertical  sections, 
will  contain  the  force  which,  under  the  assumption  made,  can 
produce  no  bending  or  compressive  stress  in  the  sections.  A 
shearing  stress  in  the  sections  is  produced  by  the  force  but  it  is 
slight  in  comparison  with  the  torsional  stress  and  is  on  this  ac- 
count also  neglected.  As  the  force  has  a  lever  arm  with  respect 
to  the  axis  of  the  coiled  bar  it  has  a  torsional  moment  with  re- 
spect to  it  and  the  corresponding  maximum  torsional  stress  in 
section  a  is,  from  equation  (1), 

„      Mtr     CDr 

o  =  -j-  =  H-T-  W 

J-P       &  IP 

in  which  C  is  the  intensity  of  the  force,  D  is  the  mean  diameter 
in  inches  of  the  helix  formed  by  the  coiled  bar,  r  is  the  distance  in 
inches  from  the  axis  of  the  bar  to  the  extreme  fibre  of  the  section, 
and  Ip  is  the  polar  moment  of  inertia  of  the  section  in  inches.4 
It  is  evident  that  the  stress  in  a  will  be  transmitted  from  section 
to  section  of  the  bar  until  the  part  resting  on  the  flat  surface  is 
reached  and,  since  these  cross-sections  are  all  alike,  equation  (5) 
gives  the  maximum  torsional  stress  S  that  occurs  in  any  part  of 
the  spring. 

117.  Fundamental  Equations  Relating  to  Helical  Springs. — 
Let  I  be  the  length  of  the  arc  through  which  the  end  of  the  ex- 
treme fibre  at  section  a  would  rotate  around  the  axis  of  the  bar  if 
straight  under  the  action  of  the  force,  and  let  L  be  the  developed 
length  of  the  portion  of  the  bar  between  section  a  and  the  point 
where  it  rises  above  the  supporting  surface  JJ,  the  end  coils  not 
being  considered.  Since  a  cross-section  of  the  end  coil,  which 
rests  on  the  supporting  surface  and  is  ground  flat  where  it  bears 
against  that  surface,  cannot  rotate  under  the  action  of  the  force, 
the  section  at  the  point  where  the  bar  rises  above  the  supporting 
surface  and  where  the  effective  part  of  the  spring  commences,  may 
be  considered  as  the  fixed  end  of  the  bar. 

The  angle  &  through  which  section  a  would  rotate  around  the 
axis  of  the  bar  if  straight  under  the  action  of  the  force  is 

e  =  l/r  radians, 


234  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

and  any  other  section  at  a  distance  x  from  the  fixed  end,  measured 
along  the  developed  length  of  the  bar,  would  rotate  around  the 
axis  of  the  bar  if  straight  through  an  angle 

I      x 
6'  =  -  X  T  radians. 

T        La 

A  section  adjoining  the  latter  and  at  a  distance  from  it  of  dx, 
measured  along  the  developed  length  of  the  bar  toward  section 
a,  would  rotate  around  the  axis  of  the  bar  if  straight  through  an 
angle 

*//      I  vx  x  +  dx      ,. 
6"  =  -X  — y —  radians, 

T  LJ 

and  relatively  to  the  latter  it  would  rotate  through  an  angle 

dd  =  -jr-  radians,  (6) 

TJU 

and  this  is  also  the  angle  through  which  the  one  section  would 
rotate  relative  to  the  other  under  the  action  of  the  force  where 
the  bar  is  coiled  into  a  helical  spring  for  the  axis  of  the  bar  may 
be  considered  straight  in  this  case  for  any  infinitesimal  length  dx. 
Let  H,  Fig.  95,  be  the  vertical  distance  from  the  center  of  this 
section  to  the  point  of  application  of  the  force  C.  Then,  since 
the  rotation  of  any  section  around  the  axis  of  the  bar  will  cause 
every  point  of  the  spring  above  it  to  move  in  the  arc  of  a  circle 
whose  center  coincides  with  that  of  the  section  or  with  the  pro- 
jection of  this  center  on  the  plane  of  rotation  of  the  point,  the 
displacement  of  the  point  of  application  due  to  the  rotation  of 
the  one  section  under  consideration  past  the  other  through  an 
angle  dB  around  the  axis  of  the  bar  is 

ds  =  \H*+  (D/2y\*de  (7) 

This  displacement  may  be  resolved  into  horizontal  and  vertical 
components  dh  and  dv,  respectively,  as  follows: 

dh  =  Hde  (8) 

and  dv  =  -5-  d&  (9) 

As  each  cross-section  rotates  around  the  axis  of  the  bar  through 
an  angle  dd  with  respect  to  that  immediately  adjoining  it  and 


COUNTER-RECOIL  SPRINGS  235 

nearer  the  supporting  surface,  the  point  of  application  of  the 
force  will  undergo  successively  the  component  displacements 
given  in  equations  (8)  and  (9),  it  being  understood  in  this  con- 
nection that  the  distance  H  is  a  variable.  If  we  consider  only 
the  horizontal  components  of  the  displacements  it  will  be  seen 
that,  as  we  pass  around  one  complete  coil  of  the  spring  from  any 
point  on  the  bar  to  a  corresponding  point  immediately  above  it, 
the  horizontal  components  neutralize  each  other  and  we  may  write 

L  Hde  =  0 


The  vertical  components  of  the  displacements,  however,  are 
seen  to  be  cumulative  so  that  the  total  displacement  of  the  point 
of  application  under  the  action  of  the  force  is  in  the  direction  of 
the  axis  of  the  spring  and  equal  to 

CLDdd_    CLDldx_Da_Dl 
Jo    ~2~~  Jo  ~27L  ==  ~26  ~  2~r 

from  which  we  may  deduce  that  the  total  displacement  of  the  point 
of  application  of  the  force,  or  the  total  compression  of  the  spring 
caused  by  the  force,  is  a  movement  in  the  direction  of  the  axis  of  the 
spring  equal  to  the  mean  radius  of  the  coil  multiplied  by  the  total 
angle,  expressed  in  radians,  through  which  the  section  at  the  free 
end  of  the  effective  portion  of  the  bar  rotates  under  the  action  of  the 
force. 

Calling  the  compression  of  the  spring  Af'A,  in  which  N'  is  the 
number  of  effective  coils  and  A  is  the  compression  per  coil,  we 
have 


and  substituting  for  l/r  its  value  from  equation  (4)  and  for  Mt 
in  the  latter  equation  its  value  CD/2  ' 


The  developed  length  of  the  effective  part  of  the  bar  is  approxi- 
mately N'irD,  and  substituting  this  value  for  L  in  equation  (11) 
and  solving  for  A  we  have  as  the  compression  per  coil 


A  =  1E 


236  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

Solving  equation  (5)  for  C  we  have 

257. 


(13) 


Equations  (11),  (12),  and  (13)  are  the  fundamental  equations 
relating  to  helical  springs.  Equations  (11)  and  (12)  give  the 
total  compression  (or  extension)  and  the  compression  (or  exten- 
sion) per  coil,  respectively,  caused  by  a  force  C;  and  equation 
(13)  gives  the  force  C  which  will  cause  a  maximum  stress  S  in 
the  fibers  of  the  bar  from  which  the  spring  is  coiled.  By  giving 
to  S  in  equation  (13)  the  maximum  value  permissible  in  con- 
nection with  the  quality  of  the  material,  the  maximum  force 
which  the  spring  can  support  without  undergoing  a  permanent 
distortion  can  be  determined.  The  compression  corresponding 
to  this  maximum  force  is  then  obtained  from  equation  (11) 
or  (12). 

Equations  (11),  (12),  and  (13)  have  been  deduced  under  the 
supposition  that  the  force  compresses  the  spring  but  they  are 
equally  applicable  when  the  force  extends  it.  In  the  latter  case 
special  means  have  to  be  provided  for  holding  the  spring  at  one 
end  and  to  enable  the  force  to  be  applied  in  the  line  of  the  axis 
of  the  spring  at  the  other  end.  Supposing  these  means  to  have 
been  provided  for  the  spring  shown  in  Fig.  95,  it  is  evident  that 
if  the  force  C  acts  upward  it  will  extend  the  spring  producing  in 
the  bar  from  which  it  is  coiled  a  torsional  stress  whose  intensity 
will  be  the  same  as  when  the  force  acts  downward.  The  angle 
l/r  through  which  section  a  rotates  under  the  action  of  the  force 
when  it  acts  upward  will  also  have  the  same  numerical  value  as 
when  the  force  acts  downward,  but  the  direction  of  rotation  of 
the  section  around  the  axis  of  the  bar  will  be  opposite  in  the  two 
cases. 

118.  Helical  Springs  Coiled  from  Bars  of  Circular  Cross- 
Section.  —  For  a  circular  cross-section  the  value  of  Ip  is  Trd*/32 
ins.4,  d  being  the  diameter  of  the  bar.  Substituting  this  value 
of  Ip  in  equation  (13)  it  becomes 

r      .3927  Sd3 
~~~ 


COUNTER-RECOIL  SPRINGS.  237 

Substituting  the  value  of  Ip  in  equations  (11)  and  (12)  they  be- 
come, respectively, 

o  /~i  mr 

(15) 

and  A  -  ^  (16) 

Equations  (14),  (15),  and  (16)  are  the  formulas  generally  used 
in  computing  the  strength  and  compression  or  extension  of 
helical  springs  coiled  from  bars  of  circular  cross-section. 

119.  Helical  Springs  Coiled  from  Bars  of  Rectangular  Cross- 
Section. —  As  first  shown  by  Saint- Venant,  an  eminent  French 
investigator,  a  plane  section  whose  axes  are  unequal  becomes  a 
warped  surface  when  subjected  to  great  torsional  strain,  and  its 
polar  moment  of  inertia  is  not  then  equal  to  the  sum  of  its  mo- 
ments of  inertia  about  two  axes  in  its  plane  perpendicular  to 
each  other  passing  through  its  center  of  gravity.  Reuleaux 
states  that  the  polar  moment  of  inertia  of  a  rectangle  when  sub- 
jected to  great  torsional  strain  is 

IP  =  Q  /to  _i_  ia\  i118-4 

3  (fi2  +  o2) 

and  that  the  distance  from  the  center  of  gravity  to  the  point  of 
the  section  most  distant  from  it  is 

hb 


in  which  h  and  6  are  the  sides  of  the  rectangle  expressed  in  inches. 
Substituting  these  expressions  for  Ip  and  r  in  equation  (13),  we 
have  _ 

r      2£fe363Vfe2+b2  =  2S  hW 

£>3  (h2  +  62)  hb  "  D  A  3  VF+T2 

and  substituting  the  expression  for  Ip  in  equations  (11)  and  (12) 
they  become,  respectively, 


/A  vx  ,1R 

A  =  x    m* 


A  A        ^rCI* 

and  A  =     A  F     X 


238  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

Equations  (17),  (18),  and  (19)  are  Reuleaux's  formulas  for 
helical  springs  coiled  from  bars  of  rectangular  cross-section. 

Equations  (14)  to  (19),  inclusive,  are  used  by  the  Ordnance 
Department,  U.  S.  Army,  in  the  design  of  counter-recoil  springs. 
Since  D  is  the  mean  diameter  of  the  helix,  the  outer  diameter  of  a 
helical  spring  coiled  from  a  bar  of  circular  cross-section  is  D  +  d 
and  its  inner  diameter  is  D  —  d.  The  outer  diameter  of  a  helical 
spring  coiled  from  a  bar  of  rectangular  cross-section  is  D  +  b 
and  its  inner  diameter  is  D  —  b,  b  being  the  side  of  the  rectangle 
that  is  perpendicular  to  the  axis  of  the  spring. 

120.  Requirements  to  be  Fulfilled  by  a  Counter-Recoil  Spring. 
—  Nomenclature.  —  In  the  design  of  a  counter-recoil  spring  the 
principal  requirements  to  be  fulfilled  are  the  following: 

(a)  The  spring  must  have  sufficient  power  to  return  the  gun 
into  battery  with  certainty  at  any  elevation  for  which  the 
carriage  is  designed. 

(6)  Its  length  when  the  gun  is  in  battery,  called  its  assembled 
height,  must  not  be  inconveniently  great. 

(c)  It  must  be  capable  of  sustaining  indefinitely  a  compression 
corresponding  to  its  assembled  height  and  of  being  re- 
peatedly compressed  through  a  further  distance  somewhat 
greater  than  the  length  of  recoil  of  the  gun,  without  suffer- 
ing permanent  deformation. 

(d)  The  outer  diameter  of  the  coiled  spring  must  not  be  in- 
conveniently great. 

(e)  Its  inner  diameter  must  be  large  enough  to  permit  the 
spring  to  pass  over  the  spring  piston-rod,  hydraulic  recoil 
cylinder,  or  other  part  on  which  the  spring  is  assembled. 

Let 

C  be  the  maximum  capacity  of  the  spring  in  pounds,  that  is, 
the  maximum  force  which  it  can  exert  or  which  can  act 
against  it  without  causing  the  maximum  stress  in  it  to 
exceed  a  permissible  amount. 

T  be  4ts  capacity  in  pounds  at  assembled  height. 

A  be  the  compression  per  coil  in  inches. 

Ar  be  the  remaining  compression  per  coil  at  assembled  height 
in  inches. 

N'  be  the  number  of  effective  coils  in  the  spring. 


COUNTER-RECOIL  SPRINGS  239 

N  be  the  total  number  of  coils  in  the  spring. 

N'A  be  the  total  compression  of  the  spring  in  inches. 

ATA,  be  the  total  remaining  compression  at  assembled  height 

in  inches. 
I   be  the  length  of  recoil  in  inches. 

A  be  the  assembled  height  of  the  spring  in  inches. 

H  be  the  solid  height  of  the  spring  in  inches. 

F  be  the  free  height  of  the  spring  in  inches. 

P  be  the  number  of  sections  of  the  spring. 

S  be  the  maximum  permissible  stress,  which  will  be  taken  as 
100000  Ibs.  per  sq.  in. 

E  be  the  torsional  modulus  of  elasticity,  equal  to  12600000 

Ibs.  per  sq.  in. 
a  be  the  maximum  angle  of  elevation  of  the  gun. 

W  be  the  weight  in  pounds  of  the  recoiling  parts. 

B  be  the  friction  in  pounds  of  the  packing  around  the  piston- 
rod  of  the  hydraulic  cylinder. 

/    be  the  coefficient  of  starting  friction. 

121.  Design  of  Counter-Recoil  Springs  Coiled  from  Bars  of 
Circular  Cross- Section.  —  In  order  that  the  spring  may  be  cap- 
able of  returning  the  gun  into  battery  with  certainty  after  it  has 
been  fired  at  the  maximum  angle  of  elevation  a,  the  force  T  which 
it  must  be  capable  of  exerting  at  assembled  height  must  be  at 
least  equal  to  the  component  of  the  weight  of  the  recoiling  parts 
parallel  to  the  surface  of  a  plane  inclined  at  an  angle  a  with  the 
horizontal  plus  the  friction  due  to  the  component  of  the  weight 
of  the  recoiling  parts  normal  to  this  plane  plus  the  friction  of  the 
packing  around  the  piston-rod,  or 

T  =  W  sin  a  +  fW  cos  a  +  B  (20) 

The  force  which  a  counter-recoil  spring  is  capable  of  sus- 
taining or  exerting  varies  directly  with  its  compression  so  that 
the  force  to  which  it  is  subjected  at  the  end  of  recoil  is  neces- 
sarily greater  than  that  to  which  it  is  subjected  at  assembled 
height;  and  by  varying  the  design  of  the  spring  the  ratio  C/T 
can  be  varied  within  wide  limits.  In  order,  however,  that  the 
assembled  height  of  the  spring  shall  be  as  small  as  possible  it  is 
necessary  that  it  shall  be  compressed  at  the  end  of  recoil  until 


240  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

the  coils  are  nearly  in  contact,  when  the  force  to  which  it  is  sub- 
jected will  be  nearly  the  maximum  which  it  is  capable  of  sustaining 
without  injury. 

The  force  T  which  the  spring  must  exert  as  its  assembled 
height  having  been  determined  from  equation  (20),  it  remains 
to  determine  the  force  C,  or  maximum  capacity  of  the  spring. 
The  movement  of  the  spring  required  between  the  loads  T  and 
C  will  be  fixed  and  somewhat  greater  than  the  length  of  recoil. 
We  will  place  it  equal  to  I  +  e  where  e  may  be  1  inch  or  more. 
With  the  fixed  movement,  fixed  load  T  at  assembled  height,  and 
fixed  value  of  D,  the  mean  diameter  of  the  coils,  we  will  now 
determine  the  relation  between  C  and  T  that  will  result  in  a 
minimum  solid  height  and  therefore  a  minimum  assembled  height 
of  the  spring. 

It  is  important  to  have  the  solid  and  assembled  heights  a 
minimum  for  the  reason  that  a  saving  in  weight  and  cost  of  spring 
cylinders,  piston  rods,  etc.,  is  thus  effected. 

From  equation  (16)  we  have  for  the  load  C, 

SN'CD* 
Ed*     ' 
and  for  the  load  T, 

„,,.       A,      SN'TD3 
N  (A  ~  Ar)  =  -ESS" 

Subtracting  the  latter  equation  from  the  former,  we  have, 


,  -. 

Considering  all  coils  effective,  we  have, 

.ZV'A,.  =  I  +  e  =  a  constant. 
N'd  =  H  =  solid  height  of  spring. 

Substituting  these  values  and  the  value  of  d5  obtained  from 
equation  (14)  in  the  above  equation,  we  obtain 


C-  T 

To  get  the  values  of  C  for  which  H  is  a  minimum  we  differ- 
entiate H  with  respect  to  C  and  place  the  differential  coefficient 


. 


COUNTER-RECOIL  SPRINGS  241 

equal  to  0.  This  gives,  finally,  neglecting  the  constant  co- 
efficient, 

JTJ        c  /q  C1^  (C1       T^       /"••& 
(In.        &/  O  O s  \\s  —  J.  )  —  os 

dc  =        (c  -  TY 

which  makes  C  =  2.5  T.  This  is  the  condition  that  should  be 
assumed  if  it  is  desired  to  keep  the  mean  diameter  of  the  coils  a 
fixed  amount. 

By  similar  reasoning  it  may  be  shown  that  if  instead  of  assum- 
ing D  constant,  we  assume  D  +  d,  or  the  outside  diameter  of  the 
spring  constant,  the  solid  height  will  be  a  minimum  when  C  =  2  T. 

This  is  the  condition  that  must  be  assumed  when  it  is  desired 
to  design  a  spring  to  work  inside  a  cylinder  whose  diameter  has 
been  fixed  by  other  considerations. 

If  we  assume  D  —  d,  or  the  inside  diameter  of  the  spring  con- 
stant we  obtain  C  =  3  T,  as  the  condition  for  minimum  solid 
height.  This  is  the  condition  that  must  be  assumed  when  it  is 
desired  to  design  a  spring  to  work  around  a  rod  whose  diameter 
has  been  fixed  by  other  considerations.* 

Assuming  the  first  of  the  above  conditions  we  have, 

C  =  2.5  T  =  2.5  (W  sin  a  +  fW  cos  a  +  B)  (21) 

Since  the  force  which  a  spring  is  capable  of  exerting  varies 
directly  as  its  compression  we  may  write 

A  :  A  -  Ar  :  :  C  :  T 
or 

Ar=A-~A  =  .6A  (22) 

In  order  that  the  spring  shall  not  be  compressed  at  the  end  of 
recoil  to  its  solid  height,  that  is,  until  one  coil  bears  solidly  against 
the  next,  the  total  remaining  compression  at  assembled  height 
]V'Ar  should  be  somewhat  greater  than  the  length  of  recoil.  The 
excess  of  AT'Ar  over  the  length  of  recoil  has  varied  somewhat  in 
different  designs  but  it  will  be  sufficient  under  ordinary  circum- 
stances to  take  it  equal  to  one  inch.  In  order,  however,  that 

Q 

*  The  above  demonstration  as  to  the  proper  ratio  =,  under  various  conditions 
was  prepared  by  Lt.  Col.  W.  H.  Tschappat,  Ord.  Dept.,  U.  S.  A. 


242  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

the  equations  may  be  of  general  application  this  excess  will  be 
represented  by  e,  and  we  may  write 

N'Ar  =  .6  N'A  =  I  +  e 
or 

XT/    _    5 

N 


3A       ' 

122.  Solid  Height  of  Spring  Column.  —  The  solid  height  of  a 
spring  is  its  length  when  it  is  compressed  until  each  coil  bears 
solidly  against  those  adjoining  it.  It  has  already  been  explained 
that  the  end  coils  of  a  counter-coil  spring  are  considered  to  be 
ineffective.  By  examination  of  Fig.  95  it  will  be  seen  that  in 
order  to  obtain  surfaces  at  the  ends  of  the  spring  which  are  truly 
perpendicular  to  its  axis  it  is  necessary,  in  addition  to  closing  the 
ends  of  the  end  coils  into  contact  with  the  adjoining  coils,  to 
grind  away  part  of  each  end  coil,  the  amount  ground  away  being 
dependent  on  the  amount  the  end  coil  is  flattened  down  and  on 
the  ratio  of  the  distance  between  the  coils  at  assembled  height 
to  the  dimension  of  the  bar  parallel  to  the  axis  of  the  spring. 
While  no  general  rule  can  be  given  to  cover  every  case,  it  will  be 
assumed  in  this  discussion  that  one-half  of  each  end  coil  is  ground 
away,  commencing  with  a  very  light  cut  at  the  section  adjoining 
the  effective  coil  and  increasing  the  cut  gradually  until  the  end 
of  the  end  coil  is  reduced  to  a  comparatively  thin  edge.  Under 
this  assumption  the  space  taken  up  by  each  end  coil  when  the 
spring  is  compressed  to  its  solid  height  is  only  about  one-half  of 
that  taken  up  by  an  effective  coil,  and  the  effect  of  the  end  coils 
on  the  solid  height  of  the  spring  may  be  obtained  by  considering 
that  only  half  a  coil  at  each  end  is  ineffective  but  that  its  volume 
is  equal  to  that  of  an  effective  half  coil. 

Because  of  the  difficulty  of  manufacturing  very  long  helical 
springs  it  is  frequently  necessary  to  make  a  counter-recoil  spring 
in  two  or  more  sections  that  are  placed  end  to  end  in  the 
spring  cylinder  with  pieces  of  metal  called  separators  between 
them. 

If  P  be  the  number  of  sections  of  the  spring,  d  the  diameter  of 
the  coiled  bar,  and  t  the  thickness  of  each  separator,  the  solid 
height  of  the  spring  column,  under  the  assumption  that  one-half 
a  coil  at  each  end  of  a  spring  section  is  ineffective  and  that  the 


COUNTER-RECOIL  SPRINGS  243 

volume  of  an  ineffective  half  coil  is  the  same  as  that  of  an  effective 
half  coil,  will  be 

H  =  (Nf  +  P)  d  +  (P  -  1)  t  (24) 

123.  Assembled  Height  of  Spring  Column.  —  Since  the  total 
remaining  compression  of  the  spring  column  at  assembled  height 
is  to  be  e  inches  greater  than  the  length  of  recoil  we  may  write 

A  =  (Nr  +  P)  d  +  (P  -  1)  t  +  I  +  e  (25) 

124.  Free   Height  of  Spring   Column.  —  The  free  height  of 
the  spring  column  is  its  solid  height  plus  the  total  compression 
which  it  is  capable  of  undergoing,  or 

F  =  (Nf  +  P)  d  +  (P  -  1)  t  +  N'A 

and  substituting  for  N'  in  the  last  term  of  this  expression  its 
value  from  equation  (23) 

F  •=  (N'  +  P)  d  +  (P  -  1)  t  +  5  (*3+  e)  (26) 

125.  Introduction  of  Values  of  Constants  in  Equations  (14) 

and  (23).  —  Replacing  S  in  equation  (14)  by  its  value,  100000 
Ibs.  per  sq.  in.,  and  solving  for  d,  we  have 

d  =  [8.46864  -  10]  CW  (27) 

Substituting  for  A  in  equation  (23)  its  value  from  equation  (16) 
and  for  E  in  the  latter  equation  its  value,  12600000  Ibs.  per  sq.  in. 

[6.41913]  (l  +  e)d« 
CD3 

the  figures  in  brackets  in  equations   (27)  and   (28)  being  the 
logarithms  of  the  numbers. 

126.  Order    of    Procedure.  —  In    designing   a   counter-recoil 
spring  column  the  length  of  recoil  and  the  value  of  C,  equation 
(21),  are  fixed  by  the  construction  of  the  carriage,  which  also  re- 
stricts within  comparatively  narrow  limits  the  outer  diameter  of 
the  spring  and,  consequently,  its  mean  diameter.    Moreover  the 
assembled  height  of  the  spring  column  must  not  be  inconven- 
iently great.     In  the  solution  of  the  problem  a  convenient  value 
of  D  is  first  selected  for  trial.    With  this  value  of  D  and  the 
fixed  value  of  C,  the  diameter  of  the  bar  from  which  the  spring 
is  to  be  coiled  is  obtained  from  equation  (27),  and  this  in  con- 


244  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

nection  with  the  assumed  value  of  D  determines  the  inner  and 
outer  diameters  of  the  spring.  If  these  are  satisfactory  the  re- 
quired number  of  effective  coils  is  next  obtained  from  equation 
(28)  after  substituting  therein  the  fixed  values  of  C  and  I,  the  de- 
sired value  of  e,  the  assumed  value  of  D,  and  the  resulting  value 
of  d.  The  number  of  effective  coils  will  determine  how  many 
spring  sections  are  required  in  the  spring  column,  and  this  hav- 
ing been  decided  and  a  suitable  value  for  the  thickness  of  the 
separators  assumed,  the  solid  height  of  the  column  is  given  by 
equation  (24),  the  assembled  height  by  equation  (25),  and  the 
free  height  by  equation  (26). 

If  the  assembled  height  thus  determined  is  too  great  another 
value  for  D  is  selected  and  the  assembled  height  again  determined. 
The  assembled  height  depends  directly  on  the  solid  height  and 
the  latter  upon  the  diameter  of  the  bar  and  the  number  of  coils. 
By  reference  to  equation  (27)  it  is  seen  that  the  value  of  d  in- 
creases directly  as  D$,  and  by  substituting  for  d  in  equation  (28) 
its  value  from  equation  (27)  it  will  be  found  that  the  value  of  N' 
varies  inversely  as  D*.  The  product  N'd,  which  is  practically 
the  solid  height,  therefore  varies  inversely  as  D*.  The  assembled 
height  for  a  given  length  of  recoil  consequently  decreases  rapidly 
with  the  increase  in  the  mean  diameter  of  the  spring. 

127.  Design  of  Counter- Recoil  Springs  Coiled  from  Bars  of 
Rectangular  Cross- Section.  —  Springs  coiled  from  bars  of  rec- 
tangular cross-section  are  capable  of  greater  compression  for  a 
given  solid  height  than  those  coiled  from  bars  of  circular  cross- 
section  and,  therefore,  when  the  former  are  used  C  is  taken  equal 
to  2  T,  whence 

C  =  2  (W  sin  a  +  fW  cos  a  +  B)  (29) 

and  N'*r  =  .5  AT' A  =  l  +  e 

or  N'  =  2(*  +  e)  (30) 

The  dimension  of  the  rectangular  cross-section  parallel  to  the 
axis  of  the  spring  being  h,  the  solid  height  of  the  spring  column, 
under  the  assumption  made  in  article  122  as  to  the  ineffective 
coils,  is 

H  =  (Nr  +  P)  h  +  (P  -  1)  t,  (31) 


COUNTER-RECOIL  SPRINGS  245 
the  assembled  height  is 

A  =  (Nf  +  P)  h  +  (P  -  1)  t  +  I  +  e,  (32) 
and  the  free  height  is 

F  =  (N'  +  P)  A  +  (P  -  1)  t  4-  2  (I  +  e)  (33) 

Let  r  equal  the  ratio  b/h,  whence  6  =  rh.    Substituting  this 

value  of  6  in  equation  (17),  replacing  S  by  its  value,  100000  Ibs. 
per  sq.  in.,  and  solving  for  h,  we  have 

h  =  [8.39203  -  10]  (1  +  r2)«  r~*  C*D*  (34) 


Substituting  for  A  in  equation  (30)  its  value  from  equation  (19) 
and  for  E  and  6  in  the  latter  equation  their  values  of  12600000 
Ibs.  per  sq.  in.  and  rh,  respectively, 

'),  (35) 


For  a  given  value  of  r  the  product  N'h,  like  the  product  N'd  for 
springs  coiled  from  bars  of  circular  cross-section,  varies  inversely 
asD*. 

By  substituting  in  equation  (35)  the  value  of  h  from  equation 
(34)  it  will  be  seen  that  N'  varies  approximately  inversely  as  r* 
and  the  product  N'h  approximately  inversely  as  r*.  The  solid 
and  assembled  heights  for  a  given  length  of  recoil  will,  therefore, 
decrease  rapidly  as  r  increases,  and  if  a  value  of  r  =  4  or  r  =  5 
be  assumed  the  assembled  height  will  be  much  less  than  it  would 
for  a  value  of  r  =  1.  When  r  =  1,  that  is,  when  the  bar  is  of 
square  cross-section,  the  power  and  compressibility  of  the  spring 
are  about  the  same  as  for  one  coiled  from  a  bar  of  circular  cross- 
section.  A  spring  coiled  from  a  bar  of  rectangular  cross-section 
with  a  value  of  r  from  4  to  5  has,  therefore,  a  very  decided  ad- 
vantage over  one  coiled  from  a  bar  of  circular  cross-section  in 
that  its  assembled  height  for  a  given  length  of  recoil  is  much  less 
than  that  of  the  latter  spring.  The  disadvantage  of  springs 
coiled  from  bars  of  rectangular  cross-section  has  been  the  diffi- 
culty of  their  manufacture  and  until  within  a  comparatively  few 
years  much  trouble  has  been  experienced  in  getting  satisfactory 
springs  of  this  type  from  manufacturers.  The  practical  limit 
for  r  in  connection  with  a  suitable  mean  diameter  of  the  spring 


246  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

seems  at  present  to  be  about  5,  this  value  being  used  in  the 
counter-recoil  springs  of  the  3-inch  field  carriage. 

The  method  of  designing  counter-recoil  springs  to  be  coiled 
from  bars  of  rectangular  cross-section  is  identical  in  principle 
with  that  described  for  springs  coiled  from  bars  of  circular  cross- 
section,  the  only  difference  being  that  a  value  for  r  must  first  be 
selected.  Ordinarily  this  value  will  be  taken  as  large  as  practi- 
cable, at  present  not  exceeding  about  5. 

128.  Telescoping  Springs.  —  To  reduce  the  assembled  height 
of  the  spring  column  when  the  length  of  recoil  is  great,  particu- 
larly if  it  is  desired  to  use  springs  coiled  from  bars  of  circular 
cross-section,  telescoping  springs  are  used,  as  in  the  case  of  the 
5-inch  barbette  carriage,  model  of  1903,  the  4.7-inch  siege  car- 
riage, model  of  1906,  the  6-inch  siege  howitzer  carriage,  model  of 
1908,  etc. 

The  principle  of  the  telescoping  spring  is  shown  in  Fig.  96.  A 
is  the  spring  cylinder  attached  to  a  non-recoiling  part  of  the 


Fig.  96. 

carriage,  P  is  the  spring  piston-rod  attached  to  the  gun,  S0  is  the 
outer  and  Si  the  inner  spring,  or  spring  column  in  case  there  is 
more  than  one  spring  section,  and  H  is  the  stirrup  connecting 
the  inner  and  outer  spring  columns.  The  stirrup  is  a  hollow 
cylinder  of  steel  with  an  inward  projecting  flange  /  at  its  rear 
end  and  an  outward  projecting  flange  /'  at  its  front  end.  When 
the  gun  recoils  it  draws  the  spring  piston-rod  with  it,  compress- 
ing the  spring  column  Si  between  the  spring  piston  p  and  the  inner 
flange  /  of  the  stirrup.  At  the  same  time  the  pressure  on  the 
flange  /  is  communicated  through  the  stirrup  and  its  outer  flange 
/'  to  the  outer  spring  column  S0,  compressing  it  also.  Owing  to 


COUNTER-RECOIL  SPRINGS  247 

the  compression  of  the  outer  spring  column  the  rear  ends  of  the 
stirrup  and  inner  spring  column  will  move  out  of  the  opening 
provided  for  the  purpose  in  the  spring  cylinder,  but  the  rear  end 
of  the  outer  column  rests  against  the  end  of  the  spring  cylinder 
and  cannot  move  during  recoil.  The  sum  of  the  compressions 
of  the  two  columns  is  equal  to  the  length  of  recoil  of  the  gun. 
It  is  evident  that  the  number  of  spring  columns  connected  by 
stirrups  can  be  increased  to  three  or  more,  if  desired,  with  con- 
sequent decrease  in  the  assembled  height  for  a  given  length  of 
recoil.  The  forces  exerted  by  the  spring  columns  must  be  equal 
to  each  other  at  all  times. 

In  a  special  case  such  as  that  of  the  5-inch  barbette  carriage, 
model  of  1903,  to  be  described  later,  it  may  not  be  advisable  to 
permit  the  rear  ends  of  the  stirrup  and  inner  spring  column  to  be 
drawn  through  an  opening  in  the  spring  cylinder  during  recoil. 
When  this  is  so  the  assembled  height  of  the  inner  column  must 
be  such  that  it  can,  when  the  gun  is  in  battery,  be  held  by  the 
stirrup  away  from  the  rear  end  of  the  spring  cylinder  at  a  distance 
at  least  equal  to  the  amount  by  which  the  outer  column  is  com- 
pressed during  recoil  plus  the  thickness  of  the  rear  flange  of  the 
stirrup,  for  otherwise  the  end  of  the  stirrup  would  strike  the  rear 
end  of  the  cylinder  during  recoil. 

129.  Design  of  Telescoping  Springs.  —  As  in  the  case  of  non- 
telescoping  springs  the  mean  diameters  of  the  springs  are  first 
selected  for  trial,  keeping  in  view  the  limitations  imposed  by  the 
design  of  the  carriage.  A  convenient  mean  diameter  for  the 
outer  spring  would  be  selected  and  the  mean  diameter  of  the 
inner  spring  made  as  great  as  possible  under  the  circumstances, 
taking  into  consideration  the  necessary  thickness  of  wall  of  the 
stirrup  and  the  clearances  required  between  it  and  the  springs. 
To  allow  for  variations  in  the  diameters  of  the  springs  during 
manufacture  and  for  the  bulging  outward  of  the  springs,  which  is 
likely  to  occur  when  they  are  compressed  nearly  to  their  solid 
height,  it  is  well  to  allow  a  clearance  of  about  .3  in.  between  the 
outer  diameter  of  the  inner  spring  and  the  inner  diameter  of  the 
stirrup  and  between  the  outer  diameter  of  the  outer  spring  and 
the  inner  diameter  of  the  spring  cylinder.  The  thickness  of  wall 
of  the  stirrup  and  the  thickness  of  its  flanges  are  determined  in 
accordance  with  the  principles  discussed  in  Chapter  IV.  The 


248  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

mean  diameters  of  the  springs  having  been  selected,  the  diame- 
ters of  the  bars  from  which  they  must  be  coiled  are  next  obtained 
from  equation  (27). 

In  the  ordinary  case  the  assembled  heights  of  the  inner  and 
outer  spring  columns  will  be  equal.  Let  the  subscript  1  be  used 
to  distinguish  the  symbols  pertaining  to  the  outer  spring  column 
and  the  subscript  2  to  distinguish  those  pertaining  to  the  inner 
column,  and  let  h  +  e/2  and  k  +  e/2  be  the  total  remaining 
compressions  at  assembled  height  of  the  outer  and  inner  spring 
columns,  respectively.  Then  we  may  write  from  equation  (25) 

(N\  +  Pi)  di  +  (Pi  -  1)  «i  +  k  +  e/2  =  (N't  +  P,)  dz  + 
(Pi  -  1)  <2  +  I*  +  e/2 

or,  since  P  and  t  will  ordinarily  be  the  same  in  the  outer  and 
inner  spring  columns, 

(N\  +  P)  dx  +  h  =  (N't  +  P)d2  +  k  (36) 

From  equation  (28),  noting  that  I  +  e  becomes  in  this  case 
Zi  +  e/2  for  the  outer  spring  column  and  ^  +  e/2  for  the  inner, 

[6.41913]  (I,  +  e/2}  dS 
CDS 

[6.41913]  ft  +  e/2)  &4 

and  Nz= 


and  substituting  these  values  in  equation  (36)  it  becomes 
[6.41913]  (h 


[[6.41918] 
We  also  have 


+  ?2  =  I  (38) 


In  equations  (37)  and  (38)  C  and  J  are  known  from  the  con- 
struction of  the  carriage,  P  and  e  are  also  known,  their  values 
having  been  decided  upon  earlier,  A  and  A  have  been  assumed 
and  di  and  dz  calculated  from  equation  (27).  The  only  unknown 
quantities,  therefore,  are  li  and  k  and  their  values  can  be  ob- 
tained by  the  solution  of  the  equations.  Having  obtained  the 


COUNTER-RECOIL  SPRINGS 


249 


values  of  h  and  ^  the  corresponding  values  of  N\  and  Af'2  are 
obtained  from  equation  (28)  and  the  solid,  assembled,  and  free 
heights  from  equations  (24),  (25),  and  (26),  respectively.  As  a 
check  on  the  accuracy  of  the  work  the  assembled  heights  should 
be  equal.  If  the  assembled  height  of  the  spring  columns  thus 
determined  is  not  satisfactory  other  values  for  DI  and  D2  would 
be  selected  and  the  assembled  height  re-determined  and  so  on. 
It  is  of  course  not  essential  that  the  assembled  heights  of  the 
inner  and  outer  columns  shall  be  equal,  and  the  total  required 
compression  may  if  desired  be  divided  between  the  columns  in 
any  arbitrary  manner.  For  a  given  length  of  recoil,  however, 
the  least  assembled  height  is  obtained  when  it  is  the  same  for 
both  columns. 

130.  Design  of  Non-Telescoping  Springs  Assembled  One 
Within  the  Other.  —  If  it  is  desired  to  decrease  the  assembled 
height  somewhat  without  increasing  the  outer  diameter  of  the 
spring,  the  amount  of  the  desired  decrease  not  being  so  con- 
siderable as  to  require  the  use  of  telescoping  springs,  it  can  be 
done  by  placing  one  spring  within  the  other  as  shown  in  Fig.  97, 


Fig.  97. 

in  which  case  each  spring  column  sustains  a  part  only  of  the  total 
load.  When  springs  are  assembled  in  this  manner  the  inner 
and  outer  springs  must  be  coiled  in  opposite  directions  so  that  if 
one  breaks  the  broken  bar  will  not  in  untwisting  be  caught  be- 
tween the  coils  of  the  other. 

In  the  equations  that  follow  the  symbols  relating  to  the  outer 
spring  column  will  be  distinguished  by  the  subscript  1  and  those 
relating  to  the  inner  column  by  the  subscript  2.  From  Fig.  97 
it  is  seen  that  the  assembled  heights  of  the  inner  and  outer  spring 
columns  must  be  equal,  and,  if  a  minimum  assembled  height  is 


250  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

desired,  the  total  remaining  compressions  at  assembled  height  and 
the  solid  heights  of  the  columns  should  also  be  equal.  Since  the 
solid  heights  are  to  be  equal  we  may  write  from  equation  (24) 

(N\  +  Pi)  di  +  (Pi  -  1)  fa  =  (N't  +  P2)  d2  +  (P2  -  1)  fa 

In  this  equation  we  may  neglect  the  terms  P\d\,  P2d2,  (Pi  —  1)  fa, 
and  (P2  —  1)  tz  for  P  may  be  taken  the  same  in  both  columns  and 
the  slight  difference  in  assembled  height  due  to  the  difference  in 
the  diameter  of  the  bars  in  the  end  coils  of  the  two  columns  can 
be  made  up  by  the  difference  in  thickness  of  the  separators  in 
the  columns,  if  any  are  used,  or  by  the  manufacturer  in  a  number 
of  ways,  by  increasing  slightly  the  number  of  coils  in  the  inner 
spring  or  by  increasing  slightly  the  pitch  of  the  coils  at  assembled 
height,  it  being  understood  that  the  manufacturer  is  not  limited 
by  the  specifications  to  the  number  of  coils  but  only  to  the  strength 
of  the  springs  at  assembled  and  solid  heights,  to  inner  and  outer 
diameters,  and  to  maximum  solid  heights  which  must  not  be 
exceeded. 
Making  these  omissions,  we  have 


and  substituting  for  N\  and  N't  their  values  from  equation  (28) 
and  for  d\  and  d2  in  the  resulting  expression  their  values  from 
equation  (27)  we  have 

[6.41913]  (fa  +  e)  \  [8.46864  -  10]  Ci*  A*}6  _ 
CiA3 

[6.41913]  (fa  +  e)  j  [8.46864  -  10]  C2*  A*|5 
C2A3 

and  reducing,  since  li  +  e  =  fa  +  e, 

Cj/Di*  =  CJ/D£  or  d/A2  =  C2/DJ  (39) 

We  also  have 

Ci  +  C2  =  C  (40) 

For  any  required  value  of  C  the  values  of  Ci  and  C2  can  be  ob- 
tained from  equations  (39)  and  (40)  after  the  values  of  A  and  Z)2 
have  been  decided  upon.  A  is  first  selected  and  A  is  made  as 
nearly  equal  to  it  as  possible  keeping  in  mind  a  required  clearance 


COUNTER-RECOIL  SPRINGS  251 

of  about  .3  in.  between  the  inner  diameter  of  the  outer  spring  and 
the  outer  diameter  of  the  inner  spring. 

If  there  are  three  or  more  springs  assembled  one  within  the 
other  we  may  write 

d/A2  =  C2/A2  =  C3/A2,  etc. 
and  Ci  +  C2  +  C3  +  etc.  =  C 

from  which  the  values  of  d,  d,  d,  etc.,  can  be  determined  after 
the  values  of  A,  A,  A,  etc.,  have  been  decided  upon. 

Having  obtained  the  values  of  Ci,  d,  etc.,  the  diameter  of  the 
bar  from  which  each  spring  is  to  be  coiled,  the  number  of  effec- 
tive coils,  the  solid  height,  assembled  height,  and  free  height  of 
each  spring  or  spring  column  can  be  determined  from  equations 
(27),  (28),  (24),  (25),  and  (26),  respectively.  Any  adjustment  of 
the  thicknesses  of  the  separators  or  other  measures  to  make  the 
assembled  heights  of  the  spring  columns  exactly  the  same  can 
now  be  decided  upon. 

131.  Measures  for  Decreasing  Assembled  Height  also  Appli- 
cable to  Springs  Coiled  from  Bars  of  Rectangular  Cross-Section.  — 
In  the  discussions  of  telescoping  springs,  and  of  non-telescoping 
springs  assembled  one  within  the  other,  it  has  been  assumed  that 
the  bars  from  which  the  springs  are  coiled  are  of  circular  cross- 
section  because  special  measures  for  decreasing  the  assembled 
heights  of  such  springs  are  more  often  necessary  than  for  springs 
coiled  from  bars  of  rectangular  cross-section.    The  principles  and 
methods  discussed  are,  however,  equally  applicable  to  springs  of 
the  latter  type. 

132.  Counter-Recoil  Springs  for  the  5-Inch  Barbette  Carriage, 
Model  of  1903.  —  The  5-inch  barbette  carriage,  model  of  1903, 
is  provided  with  two  spring  cylinders  and  two  sets  of  springs 
acting  together  on  the  gun  through  the  spring  piston-rods  and  a 
spring  yoke,  which  is  a  cross  piece  bearing  against  the  rear  of  the 
recoil-band  lug  and  carried  on  the  recoil  piston-rod.    A  spring 
piston-rod  is  fastened  to  each  end  of  the  spring  yoke.    With  this 
arrangement  each  set  of  springs  has  to  exert  but  one-half  of  the 
force  required  to  return  the  gun  into  battery  after  recoil  has 
ended. 

The  spring  cylinders  were  designed  primarily  for  springs  coiled 
from  bars  of  rectangular  cross-section  but  as  difficulty  had  been 


252  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

experienced  in  obtaining  satisfactory  springs  of  this  type  it  was 
decided  to  allow  springs  coiled  from  bars  of  circular  cross-section 
to  be  used  if  desired.  As  the  assembled  height  of  a  spring  column 
of  the  latter  springs  assembled  in  the  ordinary  way  is  greater 
than  the  length  of  the  spring  cylinder  it  was  decided  to  provide 
for  telescoping  springs;  and  further,  as  it  was  desirable  that  the 
spring  cylinder  be  capable  of  receiving  either  type  of  spring,  an 
opening  could  not  be  provided  in  the  cylinder  to  allow  the  rear 
ends  of  the  stirrup  and  inner  telescoping  spring  to  be  drawn 
through  it  during  recoil  as  is  ordinarily  the  case  when  telescoping 
springs  are  used.  It  was,  consequently,  necessary,  in  order  to 
prevent  the  stirrup  from  striking  the  rear  end  of  the  cylinder 
during  recoil,  to  shorten  it  and  the  inner  spring  column  so  as  to 
leave  a  space  between  them  and  the  rear  end  of  the  spring  cylinder 
when  the  gun  is  in  battery  somewhat  greater  than  the  amount  by 
which  the  outer  spring  is  compressed  during  recoil. 

Fig.  98  shows  a  spring  cylinder  of  this  carriage  containing 
springs  coiled  from  bars  of  rectangular  cross-section  and  also  a 
cylinder  containing  telescoping  springs  coiled  from  bars  of  cir- 
cular cross-section. 

133.  Example  1.  —  Let  it  be  required  to  design  a  counter- 
recoil  spring  coiled  from  a  bar  of  rectangular  cross-section  for 
use  in  the  5-inch  barbette  carriage,  model  of  1903.  The  follow- 
ing data  are  known  from  the  construction  of  the  carriage,  viz.: 

Weight  of  recoiling  parts  =  12632  Ibs. 

Friction  of  packing  around  recoil  piston-rod  =  220  Ibs. 

Length  of  recoil  =  13  ins. 

Maximum  angle  of  elevation  =  15°. 

Coefficient  of  starting  friction  =  .25. 

Number  of  spring  cylinders  =  2. 

Assume  values  of  e  =  1  in.,  r  =  4.25,  and  D  =  5.225  ins. 

From  equation   (29),   since  there  are  two  spring  cylinders, 
C  =  |  (12632  sin  15°  +  .25  X  12632  cos  15°  +  220)  =  6540  Ibs. 

Jy 

From  equation  (34) 

h  =  [8.39203  -  10]H  +  (4.25)2}*(4.25)-*(6540)*(5.225)*  =  .499  in. 


COUNTER-RECOIL  SPRINGS 


253 


254  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

From  equation  (35) 

_  [7.02919]  14  (.499)-          (4.25)*      _ 

6540  (5.225)3       X  1  +  (4.25)2  lls> 

On  account  of  the  number  of  coils  the  spring  will  be  divided 
into  two  sections  and  the  thickness  of  the  separator  will  be  taken 
as  .5  in.  The  solid  height  is,  then,  from  equation  (31), 

H  =  (40.07  +  2)  x  .499  +  .5  =  21.49  ins. 
The  assembled  height  is,  from  equation  (32), 
A  =  21.49  +  14  =  35.49  ins., 

and,  since  the  length  of  the  space  available  for  the  spring  column 
in  the  cylinder  is  36.75  ins.,  the  assembled  height  is  satisfactory. 
The  free  height  is,  from  equation  (33), 

F  =  21.49  +  2  x  14  =  49.49  ins. 

The  dimension  of  the  bar  perpendicular  to  the  axis  of  the 
spring  is 

b  =  rh  =  4.25  X  .499  =  2.121  ins. 

The  outer  diameter  of  the  coil  is 

D  +  b  =  5.225  +  2.121  =  7.346  ins. 

and,  since  the  inner  diameter  of  the  spring  cylinder  is  7.75  ins., 
the  clearance  is  sufficient.  The  inner  diameter  of  the  coil  is 

D  -  b  =  5.225  -  2.121  =  3.104  ins. 

and,  since  the  larger  diameter  of  the  spring  piston-rod  is  only 
1.75  ins.,  the  inner  diameter  of  the  spring  is  satisfactory. 

Compare  these  results  with  the  data  given  in  the  specifica- 
tions for  the  5-inch  barbette  carriage,  model  of  1903,  noting  that 
allowance  is  therein  made  for  a  total  remaining  compression  of 
the  spring  column  at  assembled  height  equal  to  15  ins.  In  order 
that  the  cross-section  of  the  coiled  bar  should  conform  to  a  com- 
mercial size  the  spring  just  designed  would  in  practice  be  made  of 
a  bar  having  a  cross-section  of  .5  in.  x  2.125  ins. 

134.  Example  2.  —  Let  it  be  required  to  determine  the  as- 
sembled height  of  a  column  of  counter-recoil  springs  for  the  5-inch 
barbette  carriage,  model  of  1903,  the  springs  being  coiled  from 


COUNTER-RECOIL  SPRINGS  255 

bars  of  circular  cross-section  and  having  such  an  outer  diameter 
that  they  can  be  used  in  the  existing  spring  cylinder  provided  the 
assembled  height  is  not  too  great. 

From  equation   (21),   since  there  are  two  spring  cylinders, 

C  =  ^  (12632  sin  15°  +  .25  x  12632  cos  15°  +  220)  =  8175  Ibs. 

Assuming  D  =  6.29  ins.  we  have,  from  equation  (27), 
d  =  [8.46864  -  10]  (8175)^(6.29)*  =  1.095  ins. 

The  outer  diameter  of  the  spring  is,  therefore, 
6.29  +  1.095  =  7.385  ins. 

and,  since  the  inner  diameter  of  the  cylinder  is  7.75  ins.,  the 
clearance  is  sufficient. 
From  equation  (28),  taking  e  as  1  in., 

„,      [6.41913]  14  (1.095)4      OKOr7   „ 

N'  =  -     8175  (6  2918 =  effective  coils. 

Dividing  the  spring  into  two  sections  and  taking  the  thickness  of 
the  separator  as  .5  in.,  the  assembled  height  is,  from  equation  (25), 

A  =  (25.97  +  2)  1.095  +  .5  +  14  =  45.13  ins. 

As  the  length  of  the  space  available  for  the  spring  column  in 
the  cylinder  is  only  36.75  ins.  springs  coiled  from  bars  of  circular 
cross-section  cannot  be  used  therein  unless  special  measures  are 
taken  to  diminish  their  assembled  height. 

135.  Example  3.  —  Let  it  be  required  to  determine  the  as- 
sembled height  of  a  column  of  counter-recoil  springs  for  this  car- 
riage formed  by  placing  springs  of  smaller  diameter  inside  the 
larger  springs,  the  springs  being  coiled  from  bars  of  circular 
cross-section  and  acting  directly  on  the  spring  piston. 

Assume  the  mean  diameter  of  the  outer  spring  to  be  6.45  ins., 
from  which  it  follows  that  to  allow  for  proper  clearance  between 
the  inner  and  outer  springs  the  mean  diameter  of  the  inner  should 
not  exceed  about  4.45  ins. 

From  equation  (39) 


256  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

and  since  C  =  8175  Ibs.  as  before,  we  have,  from  equation  (40), 

d  +  C2  =  1.476  d  =  C  =  8175  Ibs. 
or  Ci  =  5539  Ibs. 

From  equation  (27) 

di  =  [8.46864  -  10]  (5539)^(6.45)*  =  .969  in. 

and  from  equation  (28),  taking  e  =  1  in., 

[6.419131 14  (.969)4 

N'i  =  L    t-eon/e  AK\* =  21-80  effective  coils. 

5oo9  (o.4o)'J 

Dividing  the  spring  column  into  two  sections  and  taking  the 
thickness  of  the  separator  as  .5  in.,  the  assembled  height  of  the 
outer  column  (and  of  the  inner  column  also)  is,  from  equation  (25), 

A!  =  (21.80  +  2)  x  .969  +  .5  +  14  =  37.56  ins. 

By  comparison  with  the  assembled  height  of  a  column  of 
springs  coiled  from  bars  of  circular  cross-section  and  assembled 
in  the  ordinary  way  it  will  be  seen  that  a  material  reduction  in 
assembled  height  has  been  made,  but  this  height  is  still  greater 
than  that  of  a  column  of  springs  coiled  from  bars  of  rectangular 
cross-section  and  assembled  in  the  ordinary  way.  Although  the 
assembled  height  is  greater  than  the  length  of  the  space  available 
for  the  spring  column  in  the  cylinder,  the  difference  is  so  small 
that  by  using  springs  coiled  from  bars  of  slightly  smaller  diameters 
and  assuming  a  slightly  higher  value  for  the  permissible  torsional 
stress,  springs  could  be  obtained  that,  when  assembled  in  this 
way,  would  fit  in  the  spring  cylinder  of  this  carriage  and  function 
satisfactorily.  Such  springs  would,  however,  require  greater  care 
in  their  production  by  the  manufacturer. 

136.  Example  4.  —  Let  it  be  required  to  design  telescoping 
springs  for  this  carriage  to  be  coiled  from  bars  of  circular  cross- 
section,  and  with  the  condition  that  the  spring  cylinder  must  not 
be  altered  in  any  respect  to  the  end  that  either  springs  coiled 
from  bars  of  rectangular  cross-section  or  telescoping  springs 
coiled  from  bars  of  circular  cross-section  may  be  used  therein  at 
will. 

Since  the  spring  cylinder  must  not  be  altered,  provision  cannot 
be  made  for  drawing  the  rear  ends  of  the  stirrup  and  the  inner 


COUNTER-RECOIL  SPRINGS  257 

spring  column  through  an  opening  in  the  cylinder  during  recoil, 
and  a  space  must  be  left  between  the  rear  end  of  the  stirrup  and 
the  rear  end  of  the  cylinder,  when  the  gun  is  in  battery,  that  is 
at  least  equal  to  the  total  remaining  compression  of  the  outer 
spring  column  at  assembled  height.  As  the  inner  spring  is  smaller 
and,  therefore,  less  expensive  than  the  outer,  the  inner  spring 
column  will  be  given  the  greatest  possible  assembled  height  con- 
sistent with  the  conditions  of  the  problem. 

Let  the  total  remaining  compression  at  assembled  height  of  the 
outer  spring  column  be  li  +  .5  and  the  total  remaining  com- 
pression at  assembled  height  of  the  inner  spring  column  be  k  +  .5, 
and  let 

li  +  .5  +  k  +  .5  =  I  +  1  =  14  ins. 

Then  the  assembled  height  of  the  inner  spring  column  is,  from 
equation  (25), 

A2  =  (N't  +  P2)  ^  +  (Pt  -  1)  <2  +  Z>  +  .5 

and  this  height  plus  li  +  .5  plus  the  thickness  of  the  flange  at 
the  rear  end  of  the  stirrup  must  equal  the  length  of  the  space 
available  for  the  spring  column  in  the  cylinder,  which  is  36.75 
ins.  Taking  the  thickness  of  the  flanges  of  the  stirrup  as  .75  in., 
we  may  write 

(N't  +  P«)  dz  +  (Pi  -  1)  *2  +  k  +  .5  +  It  +  .5  +  .75  =  36.75  ins. 
or 

(N't  +  P2)  dz  +  (Pi  -  1)  fe  =  36.75  -  14  -  .75  =  22  ins. 

which  expresses  the  fact  that  the  solid  height  of  the  longest 
inner  spring  column  that  can  be  used  in  the  cylinder  under  the 
assumed  conditions  is  22  ins. 

The  mean  diameter  of  the  outer  springs  and  the  diameter  of 
the  bars  from  which  they  are  coiled  will  be  taken  as  6.29  ins. 
and  1.095  ins.,  respectively,  as  before;  and  to  allow  for  a  proper 
thickness  of  the  wall  of  the  stirrup  and  for  the  proper  clearances 
between  the  springs  and  the  stirrup,  the  mean  diameter  of  the 
inner  springs  will  be  taken  as  3.4  ins.,  which  will  also  make  the 
inner  diameter  of  these  springs  sufficiently  large  to  enable  them 
to  pass  easily  over  the  spring  piston-rod. 


258  STRESSES  IN  GUNS  AND  GUN  CARRIAGES 

From  equation  (27) 

<*2  =  [8.46864  -  10]  (8175)*  (3.4)*  =  .891  in. 
and  taking  P2  =  Pi  =  2  and  tz  =  ti  —  .5  we  have 
(N't  +  2)  x  .891  +  .5  =  22  ins. 
or  N't  =  22.13  effective  coils. 

From  equation  (28) 

,  -      22.13  x  8175  x  (3.4)3 
[6.41913]  (.891)' 

The  amount  of  compression  to  be  provided  by  the  outer  spring 
column  is,  therefore, 

14  -  4.298  =  ^  +  .5  =  9.702  ins. 

and  the  number  of  effective  coils  in  the  outer  column  is,  from 
equation  (28), 

„,       [6.41913]  (9.702)  (1.095)* 
8175  (6.29)3 

The  assembled  height  of  the  outer  column  is,  therefore,  from 
equation  (25), 

A!  =  20  x  1.095  +  .5  +  9.702  =  32.10  ins. 

and  since  this  is  less  than  the  length  of  the  available  space  in 
the  spring  cylinder  diminished  by  the  thickness  of  the  front 
flange  of  the  stirrup,  equal  to  36.75  —  .75  =  36  ins.,  the  design 
of  telescoping  springs  just  made  is  satisfactory  for  use  in  the 
spring  cylinder  of  the  5-inch  barbette  carriage,  model  of  1903, 
designed  primarily  for  counter-recoil  springs  coiled  from  bars  of 
rectangular  cross-section. 

137.  Example  5.  —  Let  it  be  required  to  determine  the  as- 
sembled height  of  a  set  of  telescoping  springs  for  this  carriage 
under  the  assumptions  that  the  assembled  heights  of  the  inner 
and  outer  columns  are  equal,  that  the  rear  ends  of  the  stirrup 
and  inner  spring  column  may  be  drawn  through  an  opening  in 
the  spring  cylinder  during  recoil,  and  that  the  diameter  of  the 
cylinder  is  the  same  as  at  present. 


• 

COUNTER-RECOIL  SPRINGS  259 

The  values  of  D  and  d  will  be  taken  the  same  as  for  the  tele- 
scoping springs  already  designed  which  are  as  follows: 

A  =  6.29  ins.  dt  =  1.095  ins. 

A  =  3.40  ins.  d^  =    .891  in. 

Let  the  total  remaining  compression  at  assembled  height  of 
the  outer  spring  column  be  Zi  -f-  .5  and  of  the  inner  spring  column 
k  +  .5,  and  let  li  +  k  =  13  ins.,  the  length  of  recoil  required. 
Then  from  equation  (37),  assuming  P  =  2, 

3.0311  h  +  3.2056  =  5.5878  Z«  +  4.076 

and  substituting  in  this  equation  for  k  its  value  13  —  lit  and 
solving  for  lt  we  have 

li  =  8.53  ins.  and  ^  =  4.47  ins. 
From  equation  (28) 

^-^war* -»»--•••* 

and 

,  _  [6.41913]  (4.97)  (.891)*  _  2g  59   ff    ti         ^ 
8175  (3.40)3 

From  equation  (25) 

Ai  =  18.75  x  1.095  +  .5  +  9.03  =  30.06  ins. 
and 

A2  =  27.59  x  .891  +  .5  +  4.97  =  30.05  ins. 


ENGINKWNH  BUREAU 
CANNON 


ENGINEERING  BUREAU 

CANNON   SECTION 


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